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Measuring what cannot be reached
A STORY
The tank on the roof
The water tank on the school roof has a leak. The plumber has to bring a ladder, so he asks how tall the tank is. Nobody knows.
After class, Nila and Dhruv go up to the roof. The tank stands at the far end, high on four tall legs. It is too high to climb, and no tape reaches up there.
Dhruv has made a sighting tool from two strips of wood. He holds one end steady. Nila looks along its top edge at the top of the tank. The tool gives the angle: $30^\circ$ above the level of her eye.
“That is one number,” Dhruv says. “We still need the distance to the tank.”
They count their steps across the roof to the tank’s legs. It is about $20$ m.
“One angle and one distance,” Nila says. “Can that really give the height?”
It can. How would you get the height from just those two numbers?
A tower too tall to climb. A river too wide to cross. A kite string too far to reach. Each one still yields to arithmetic. We never touch the object itself.
Stand at a known distance from it. Look up or down at a measured angle. The tower or the river becomes one right triangle. You already know how to solve it. The tool is the trigonometric ratio from the last chapter: $\sin$, $\cos$ and $\tan$. Now it aims at a real angle and a real distance, not an abstract one.
Nothing new is learned about $\sin$, $\cos$ or $\tan$ here. Only the aim changes.
Your turn: stand $14$ m from a chimney’s foot and look up at $30^\circ$ to its top. Name the triangle’s three sides. (Answer: the ground distance, the chimney’s height, and the line of sight joining them.)
The unknown lies flat across the water, not standing in the air: the side next to the one measurable angle.
A SCHOLAR INDIA REMEMBERS
Brahmagupta worked in a school of astronomy that predicted eclipses with shadow geometry. Nobody could climb up to the sun or the moon. The measuring was done from the ground, with lengths and angles. You will work the same way with a tower. Measure a length and an angle from the ground. Then a right triangle gives you the height.
Brahmagupta · the brahminy kite
Check yourself
This chapter finds the height of a tower or the width of a river without touching either. What two things does the method actually need?
two distances, never an angle
two angles, never a distance
the answer itself, checked afterward
an angle and a distance
Check your answer
two distances, never an angle — the method always starts from one measured angle; two distances alone fix no triangle without it
two angles, never a distance — an angle alone has no scale; one accessible distance is what turns the angle into an actual height
the answer itself, checked afterward — the height is derived from the angle and distance; nothing here starts from a guessed answer
✓ an angle and a distance — (D) The method needs exactly one angle — of elevation or depression — and one accessible distance; the right triangle they form gives the unknown side.
Why does measuring an angle and a distance give the height of something you cannot reach?
they are two parts of one right triangle
the angle alone fixes the height, regardless of distance
the distance alone fixes the height, regardless of angle
a second, unmeasured distance is secretly compared
Check your answer
✓ they are two parts of one right triangle — (A) The angle, the known distance and the unknown height are three parts of one right triangle, so fixing two of them fixes the third.
the angle alone fixes the height, regardless of distance — an angle by itself carries no length information; the distance is what scales it into an actual height
the distance alone fixes the height, regardless of angle — the same distance fits a short building at a shallow angle and a tall one at a steep angle; the angle is what tells them apart
a second, unmeasured distance is secretly compared — only one distance is measured; the height is the triangle’s other side, not a second distance
A river’s width cannot be measured by wading across it. An observer measures $50 m$ along the bank and a $30^\circ$ angle to a point across the river. What replaces the impossible direct measurement?
estimating the width by eye, then rounding
measuring a second distance along the far bank
assuming the river has a standard width for its region
computing $50 \cdot \tan 30^\circ$
Check your answer
estimating the width by eye, then rounding — the method replaces guessing entirely — the angle and distance, run through a ratio, give an exact computed width
measuring a second distance along the far bank — only one accessible distance and one angle are needed; a second measurement across the river plays no part in this method
assuming the river has a standard width for its region — the method computes this river’s own width from this angle and distance — it never falls back on an assumed regional average
✓ computing $50 \cdot \tan 30^\circ$ — (D) The computation $50 \cdot \tan 30^\circ$, from the measured angle and distance, replaces the direct measurement that is impossible here.
You cannot climb the tower or cross the river. Stand where you are, and measure one angle. Try each check below.
Alternate angles (Class 8, Ganita Prakash Part 1, page 99). Here: the top’s depression angle equals the bottom’s elevation angle. Check: Angles on a line: $180^\circ - 125^\circ = 55^\circ$, its alternate angle too.
Angle sum of a triangle (Class 8, Ganita Prakash Part 1, page 94). Here: a right triangle’s other two angles add to $90^\circ$, so elevation is acute. Check: A right triangle with angles $90^\circ$ and $60^\circ$ has a third angle of $30^\circ$.
Multiplying both sides (Class 8, Ganita Prakash Part 1, page 170). Here: isolating $l$ in $\sin 30^\circ = 8/l$ multiplies both sides by $l$. Check: $x/4 = 3$ gives $x = 12$, multiplying both sides by $4$.
Solving a linear equation (Class 9, Ganita Manjari, page 20). Here: the flagpole’s height is the $x$ found by solving its equation. Check: $5x = 20$ gives $x = 4$.
Irrational numbers (Class 9, Ganita Manjari, page 53). Here: the tower’s height, $12 \cdot \sqrt{3}$ m, is exactly such a number. Check: $12 \sqrt{3}$ is about $20.8$ m, since $\sqrt{3}$ is about $1.732$.
If any of these felt new, read the page named before going on.
Every heights-and-distances problem starts the same way. Name one line precisely. The line of sight runs from the observer’s eye to the exact point you are looking at.
Nothing else counts as the line of sight. Not a line to the object’s base. Not a line to some other point on it. Look at the top of a tower. You see the line of sight end there, nowhere else on it.
Every angle of elevation or depression is measured from this line. Get its endpoint wrong, and your whole triangle goes wrong with it.
Your turn: someone looks at a bird resting on a wire above them. Name the two points the line of sight joins. (Answer: the observer’s eye and the bird, since a line of sight runs from the eye to the object seen.)
One ray, in teal, reaches the named point; the other, crossed out in red, wrongly points at the object’s base.
Check yourself
The line of sight, in this chapter’s diagrams, is
the line from the eye to the point viewed
the horizontal line at the observer’s eye
the vertical line up the tower or pole
the ground between the observer and the object
Check your answer
✓ the line from the eye to the point viewed — (A) The line of sight is the straight line joining the observer’s eye to the point being viewed.
the horizontal line at the observer’s eye — the horizontal at eye level is the reference the angle is measured from; the line of sight is the slanted line to the object itself
the vertical line up the tower or pole — the tower’s own vertical side is one leg of the triangle; the line of sight is the separate line from the eye to the top
the ground between the observer and the object — the ground distance is a length you measure; the line of sight is the line of viewing itself, usually the triangle’s slanted side
In the triangle drawn for a heights problem, why is the line of sight usually the slanted (hypotenuse) side, not one of the two legs?
it is always the shortest of the three sides
it lies flat along the horizontal, like the ground
it runs straight from eye to object
it always equals the vertical side in length
Check your answer
it is always the shortest of the three sides — the hypotenuse is the longest side of a right triangle; the line of sight is never the shortest one
it lies flat along the horizontal, like the ground — the horizontal is its own leg of the triangle; the line of sight tilts away from it by the marked angle
✓ it runs straight from eye to object — (C) It runs directly from the observer’s eye to the object, cutting across the triangle’s right angle, which is exactly what a hypotenuse does.
it always equals the vertical side in length — the line of sight and the vertical side are related through the trig ratio of the marked angle, not through being equal lengths
A student marks the line of sight as the same line as the horizontal at the observer’s eye. What is wrong with that?
nothing — the two lines are the same
the horizontal should instead be drawn vertical
the observer’s eye should be moved to ground level first
it is the sighting line, not the reference line
Check your answer
nothing — the two lines are the same — if the two lines were the same, every angle of elevation or depression would be zero — the chapter would have nothing left to measure
the horizontal should instead be drawn vertical — the horizontal is correctly horizontal; the error is treating the sighting line as if it were that same reference line
the observer’s eye should be moved to ground level first — nothing requires moving the eye; the line of sight simply starts where the eye actually is
✓ it is the sighting line, not the reference line — (D) The line of sight is the sighting line to the object; the horizontal is only the reference line the angle is measured from.
Look up at a point above eye level. The angle between your line of sight and the horizontal is the angle of elevation.
That angle is always measured from the horizontal. It is never measured from the vertical pole, tower or cliff face nearby. Look straight ahead, and you trace the horizontal. Tilt your head up to the object, and you trace the line of sight instead. The angle of elevation is the gap between the two.
Picture a kite above the horizontal, a plane overhead, the top of a building seen from the street. Each one gives an angle of elevation. Each is read off the same horizontal.
Your turn: someone on the ground looks up at a lit window near a building’s top. Name the angle this makes with the horizontal. (Answer: an angle of elevation, since the window sits above eye level.)
A line of sight joins two observers; each measures from their own horizontal, so elevation at the bottom equals depression at the top.
A SCHOLAR INDIA REMEMBERS
Every angle here is measured from something fixed. What is it? It is the level line through your eye. Look straight ahead. That line is $0^\circ$. Now raise your eyes to the top of the tree. The turn from the level line to the top is the angle of elevation.
Brahmagupta · the brahminy kite
Check yourself
The angle of elevation is the angle between the line of sight and
the vertical, when looking upward
the horizontal, when looking upward
the horizontal, when looking downward
the ground, at the foot of the object
Check your answer
the vertical, when looking upward — elevation is always read from the horizontal at eye level, never from the vertical pole or tower
✓ the horizontal, when looking upward — (B) The angle of elevation is the angle between the line of sight and the horizontal, formed when the observer looks upward.
the horizontal, when looking downward — looking downward at a point below eye level is depression; elevation is the upward case
the ground, at the foot of the object — the reference horizontal sits at the observer’s own eye, not at the ground below the object being viewed
An observer looks up at the top of a tower. Which angle does this line of sight make?
an angle of depression
an angle of elevation
a right angle
an alternate angle
Check your answer
an angle of depression — depression names the downward case, at a point below the observer’s eye; looking up at a tower’s top is elevation
✓ an angle of elevation — (B) Looking upward at a point above eye level makes an angle of elevation.
a right angle — the 90-degree angle is where the tower meets the ground, not the angle the line of sight makes with the horizontal
an alternate angle — alternate angle is a relationship between two angles at two different points, not the name for a single sighting angle
An observer stays the same accessible distance from a tower, but the tower is rebuilt taller. What happens to the angle of elevation measured from that same spot?
it decreases
it stays exactly the same
it depends only on the observer’s own height
it increases
Check your answer
it decreases — a taller tower at the same distance is seen along a steeper line of sight, so the angle grows, not shrinks
it stays exactly the same — the angle depends on the ratio of height to distance; changing the height without changing the distance changes the angle too
it depends only on the observer’s own height — the observer’s own height shifts where eye level sits; it does not decide how the angle responds to the tower growing taller
✓ it increases — (D) A taller tower seen from the same distance is sighted along a steeper line, so the angle of elevation increases.
Look down at a point below eye level. The angle between your line of sight and the horizontal is the angle of depression.
Like elevation, this angle is measured from the horizontal at the observer’s eye. It is never measured from the cliff, tower or wall the observer stands on. Picture a boat seen from a cliff top. Picture a car seen from a tall building’s roof. Both give an angle of depression, off that same horizontal.
The only difference between depression and elevation is which way you look, up or down, from that same horizontal.
Your turn: a person on a balcony looks down at a car parked below. Name the angle this makes with the horizontal. (Answer: an angle of depression, since the car sits below eye level.)
A line of sight from a cliff-top observer down to a boat gives the angle of depression, measured from the horizontal.
Check yourself
The angle of depression is the angle between the line of sight and
the vertical, when looking downward
the horizontal, when looking downward
the horizontal, when looking upward
the ground, at the object being viewed
Check your answer
the vertical, when looking downward — depression, like elevation, is always read from the horizontal at eye level, never from the vertical
✓ the horizontal, when looking downward — (B) The angle of depression is the angle between the line of sight and the horizontal, formed when the observer looks downward.
the horizontal, when looking upward — looking upward at a point above eye level is elevation; depression is the downward case
the ground, at the object being viewed — the reference horizontal is drawn at the observer’s eye; the object being viewed can be anywhere below it
From the top of a lighthouse, a lookout spots a boat on the sea below. The angle this line of sight makes with the horizontal at the lookout’s eye is
an angle of elevation
an angle of depression
the boat’s own angle of tilt
a right angle, since the lighthouse is vertical
Check your answer
an angle of elevation — elevation names the upward case; a boat seen below eye level gives an angle of depression instead
✓ an angle of depression — (B) A boat seen below the lookout’s eye level, along the line of sight, gives an angle of depression.
the boat’s own angle of tilt — the angle belongs to the line of sight from the lookout’s eye, not to any tilt of the boat itself
a right angle, since the lighthouse is vertical — the lighthouse’s verticality gives the triangle’s 90-degree corner at its foot; the sighting angle at the top is a separate, smaller angle
A lookout climbs higher up the same lighthouse before spotting the same boat at the same horizontal distance. What happens to the angle of depression?
it decreases
it stays exactly the same
it becomes an angle of elevation instead
it increases
Check your answer
it decreases — a greater height over the same distance is a steeper line of sight, so the angle of depression grows, not shrinks
it stays exactly the same — height changed while distance did not, so the ratio that fixes the angle changed with it
it becomes an angle of elevation instead — looking down at the boat is still depression, whatever the angle’s size; it does not switch category by growing
✓ it increases — (D) A greater height over the same horizontal distance is a steeper line of sight, so the angle of depression increases.
Elevation and depression are not two different kinds of angle. They are the same kind. We read both off the same reference line: the horizontal at the observer’s eye.
Look upward to a point above that line, and you call it an angle of elevation. Look downward to a point below it, and you call it depression instead. The horizontal itself never changes. Only the direction of the line of sight does.
Two different observers looking at two different objects still each use their own horizontal. The reference line sits at each observer’s own eye. It is never borrowed from the other observer’s line of sight.
Your turn: one person on the ground looks up at a plane. A passenger inside the plane looks down at that person. Name the angle each one measures. (Answer: elevation for the one on the ground, depression for the one in the plane, both from their own horizontal.)
Check yourself
Elevation and depression are the same kind of angle because both are measured from
the vertical pole or tower
the ground at the object’s foot
the horizontal at the observer’s eye
whichever line makes the angle come out smaller
Check your answer
the vertical pole or tower — the one line both angles share is the horizontal at eye level; the vertical belongs to the object, not to the reference
the ground at the object’s foot — the reference horizontal travels with the observer’s eye; it is not fixed to the ground under the object
✓ the horizontal at the observer’s eye — (C) Both angles are measured from the same horizontal line, drawn at the observer’s own eye level.
whichever line makes the angle come out smaller — the horizontal reference is fixed by where the observer’s eye is; it is never picked afterward to shrink the angle
If the only difference between elevation and depression is the direction of looking, what stays exactly the same between them?
the horizontal reference line itself
the size of the angle
which object is being viewed
the observer’s own height above the ground
Check your answer
✓ the horizontal reference line itself — (A) Only the horizontal reference line stays fixed; the direction of looking is the one thing that changes between elevation and depression.
the size of the angle — nothing forces the two angles to match in size; only the reference line they are both measured from stays fixed
which object is being viewed — the observer can look up at one object and down at another entirely; only the horizontal reference is shared
the observer’s own height above the ground — the observer’s height is a separate piece of information; the shared thing here is the reference line itself
A student writes: “elevation is measured from the horizontal, but depression is measured from the vertical.” What is wrong with this?
nothing — the two really do use different references
elevation should instead use the vertical
both should be measured from the ground
both are measured from the same horizontal line
Check your answer
nothing — the two really do use different references — both angles are defined from the horizontal at eye level; there is no separate vertical reference for depression
elevation should instead use the vertical — elevation is correctly measured from the horizontal already; the mistake is inventing a different reference for depression
both should be measured from the ground — neither angle is measured from the ground; both share the horizontal drawn at the observer’s own eye
✓ both are measured from the same horizontal line — (D) Both elevation and depression are measured from the same horizontal line at the observer’s eye — there is no separate vertical reference.
One equality follows directly from how elevation and depression are defined. The angle of depression at a high point equals the angle of elevation at the low point below it.
The two horizontals are parallel. One sits at the high point, one at the low point. Both are simply horizontal. The line of sight joins the two points. It cuts across both horizontals, so it acts as a transversal. A transversal crossing two parallel lines makes equal alternate angles. The angle of depression at the top and the angle of elevation at the bottom are the same angle, named from opposite ends.
You use this equality to move a given angle down. It goes from the top of a tall figure to the base of the right triangle we are actually solving. Stand at the top of a cliff and look down at a boat: the angle of depression is $30^\circ$. The boat’s own angle of elevation, looking up at the cliff top, is $30^\circ$ too.
Your turn: from a watchtower, the angle of depression to a hiker below is $45^\circ$. Name the angle of elevation from the hiker up to the watchtower. (Answer: $45^\circ$, since the two horizontals are parallel, so the two angles are alternate angles and equal.)
Alternate angles carry the given angle from the observer to the far end of the base, where the working needs it.
Check yourself
The angle of depression measured from a tall building’s top, looking at a point on the ground, equals
half the angle of elevation measured from that ground point
the angle of elevation measured from that ground point
the complement of the elevation angle from that ground point
unrelated to the elevation angle from that ground point
Check your answer
half the angle of elevation measured from that ground point — alternate angles are exactly equal; there is no halving or scaling between the top view and the ground view
✓ the angle of elevation measured from that ground point — (B) The depression angle at the top and the elevation angle at that ground point are alternate angles on a transversal cutting two parallel horizontals, so they are equal.
the complement of the elevation angle from that ground point — the two angles are equal, not complementary; treating them as summing to $90^\circ$ would double-count the same triangle wrongly
unrelated to the elevation angle from that ground point — the two horizontals are parallel and the line of sight is a transversal, so the two angles are always related — equal, in fact
Why are the depression angle at the top and the elevation angle at the ground equal, rather than just related somehow?
both angles are measured against the same vertical tower
the triangle happens to be isosceles
parallel horizontals cut by one transversal
the two points happen to be the same distance from the tower
Check your answer
both angles are measured against the same vertical tower — both angles are read from a horizontal line, one at the top and one at the ground; the tower’s vertical side is not what makes them equal
the triangle happens to be isosceles — no isosceles assumption is needed — the equality comes from parallel horizontals cut by one transversal, whatever the triangle’s proportions
✓ parallel horizontals cut by one transversal — (C) The two horizontals, one at the top and one at the ground, are parallel, and the line of sight crossing both is a transversal — that makes the two angles equal alternate angles.
the two points happen to be the same distance from the tower — the distance from the tower’s foot plays no role in the equality; it comes purely from the two horizontals being parallel
From the top of a tower, the angle of depression of a car on the road is $40^\circ$. A student says the angle of elevation of the tower’s top, measured from the car, must be $50^\circ$. What is wrong?
nothing — $50^\circ$ is correct, since the angles add to $90^\circ$
the elevation angle needs the tower’s height to be found at all
$50^\circ$ is right only if the car sits directly below the tower
the elevation angle from the car is also $40^\circ$
Check your answer
nothing — $50^\circ$ is correct, since the angles add to $90^\circ$ — the two horizontals are parallel, so the angles are equal alternate angles, $40^\circ$ each — not complementary angles adding to $90^\circ$
the elevation angle needs the tower’s height to be found at all — alternate-angle equality needs no height at all — the depression angle at the top already fixes the elevation angle at the ground
$50^\circ$ is right only if the car sits directly below the tower — the equal-angle result holds for the car at any point on that road, not only directly below the tower
✓ the elevation angle from the car is also $40^\circ$ — (D) The two horizontals are parallel, so the depression and elevation angles are equal alternate angles — the elevation angle from the car is also $40^\circ$, not $50^\circ$.
One problem gives the angle of depression from a tall building to a shorter one; another gives two angles of elevation from one ground point to two heights on a pole. What do these two problems have in common?
both are solved using only the sine ratio
both use two right triangles sharing one side
both require an angle of depression to be given
both give the answer directly from one trigonometric ratio
Check your answer
both are solved using only the sine ratio — neither problem is tied to sine specifically; the ratio chosen depends on which sides are known and wanted in each triangle
✓ both use two right triangles sharing one side — (B) Both problems are solved with two right triangles that share one side, giving two trigonometric equations solved together.
both require an angle of depression to be given — the pole problem uses two angles of elevation and no depression angle at all; depression is not a requirement here
both give the answer directly from one trigonometric ratio — a single ratio is enough only when one triangle carries all the information; both examples here need two
Every heights-and-distances problem starts with one right triangle. It has three sides. A vertical side. A horizontal side. The line of sight joins them as the third side.
Let us draw the triangle together. We mark the right angle first, where the vertical side meets the horizontal side. That is where the tower, pole or building meets level ground. Next we mark the given angle, elevation or depression, at the observer. Then we label the side the problem gives, and the side it asks for.
Label opposite, adjacent and hypotenuse relative to the marked angle, never to the shape of the triangle. The vertical side is opposite the angle in some problems. It is adjacent to it in others. The marked angle decides the labels every time.
Your turn: stand $9$ m from a pole’s foot and look up at $60^\circ$ to its top. Find the pole’s height. (Answer: $9 \cdot \sqrt{3}$ m, since $\tan 60^\circ = h/9$ gives $h = 9 \cdot \sqrt{3}$.)
The right angle sits where the vertical meets the horizontal, and the marked angle at the observer names opposite, adjacent and hypotenuse.
Check yourself
Every heights-and-distances triangle in this chapter has three sides: a vertical side, a horizontal side, and
a second vertical side
the angle of elevation itself
the line of sight
a line joining the two given distances
Check your answer
a second vertical side — the triangle has one vertical leg and one horizontal leg; a second vertical side would make it a different shape entirely
the angle of elevation itself — the angle is marked at a corner of the triangle; it is not one of the three sides
✓ the line of sight — (C) The third side is the line of sight, joining the eye to the point being viewed.
a line joining the two given distances — the basic setup has exactly three sides; a fourth connecting line is not part of this triangle
In this triangle, where is the right angle marked?
where the line of sight meets the horizontal side
where the vertical side meets the horizontal side
where the line of sight meets the vertical side
wherever makes the two legs come out equal
Check your answer
where the line of sight meets the horizontal side — the line of sight meets the horizontal at the observer’s corner, where the given angle is marked — not the right angle
✓ where the vertical side meets the horizontal side — (B) The right angle sits where the vertical side meets the horizontal side, at the foot of the object.
where the line of sight meets the vertical side — the line of sight meets the vertical side at the object’s top; no right angle is marked there
wherever makes the two legs come out equal — the right angle sits where the vertical meets the horizontal, by the setup itself — not wherever would make the legs equal
A problem gives the height of a pole and the angle of elevation to its top, and asks for the distance to its foot. Before choosing a ratio, what must the triangle diagram show correctly?
the exact scale of the drawing, in centimetres
the compass direction the observer is facing
which side is known, which is wanted
the exact colour used for the line of sight
Check your answer
the exact scale of the drawing, in centimetres — the diagram only needs correct labels and a marked angle; it is never a scale drawing measured out in centimetres
the compass direction the observer is facing — compass direction plays no role here; the triangle only needs its sides and angle labelled correctly
✓ which side is known, which is wanted — (C) The diagram must correctly show which side is the known height and which side is the wanted distance, before any ratio is chosen.
the exact colour used for the line of sight — how the line of sight is drawn or coloured makes no difference; only which side is known and which is wanted matters
An observer rarely stands with an eye exactly at ground level. Say the observer has some height of their own. Then the triangle’s vertical side runs from EYE level, not ground level, up to the point being viewed.
Solve that triangle yourself, and we get only the part of the height above the observer’s eye. We still have to add the observer’s own height back afterward. Only then do we reach the object’s full height above the ground.
Folding the observer’s height in too early changes what the triangle’s vertical side measures. Take an observer $1.5$ m tall who finds a triangle side of $18.5$ m up to a chimney top. The chimney’s height is $18.5 + 1.5 = 20$ m, not $18.5$ m.
Your turn: an observer $1.4$ m tall stands $10$ m from a pole. The angle of elevation to its top is $45^\circ$. Find the pole’s full height. (Answer: $11.4$ m, since $\tan 45^\circ = 1$ gives a $10$ m rise above eye level, plus the observer’s $1.4$ m.)
The triangle runs only from eye level to the top; the observer’s own height, tied to it, is added back afterward.
A SCHOLAR INDIA REMEMBERS
Your answer says the tower is $30$ m tall. The builder says $31.5$ m. Who is wrong? Check your own working first. Did you measure the angle from your eye, $1.5$ m above the ground? Then your triangle starts $1.5$ m up. Add it on: $30 + 1.5 = 31.5$. Now your answer matches the tower.
Brahmagupta · the brahminy kite
Check yourself
When an observer standing on top of something looks at a distant object, the triangle’s vertical side should be measured from
the ground the observer stands on
the midpoint between the observer’s eye and feet
the observer’s eye level, not the ground
the top of whatever the observer is standing on
Check your answer
the ground the observer stands on — starting from the ground under the observer skips the observer’s own height, which still has to be added to reach the object’s true height
the midpoint between the observer’s eye and feet — no averaging is used; the triangle’s vertical side runs from the actual eye level, and the observer’s full height is added separately
✓ the observer’s eye level, not the ground — (C) The vertical side runs from the observer’s actual eye level, not from the ground below.
the top of whatever the observer is standing on — the surface the observer stands on and the observer’s eye level are not the same point; the gap between them is the observer’s own height
After solving the triangle for the vertical distance from eye level up to the object, why must the observer’s own height still be added?
to correct for a rounding error in the trig ratio
because the angle was measured incorrectly
to make the final answer come out to a whole number
the triangle stops short, at eye level
Check your answer
to correct for a rounding error in the trig ratio — the observer’s height is a real distance, not a rounding fix — it is the gap the triangle skipped by starting at eye level
because the angle was measured incorrectly — nothing is wrong with the angle itself; the triangle it produces simply stops at eye level, short of the ground
to make the final answer come out to a whole number — the height is added because it is a real distance the triangle omitted, not to tidy the final number
✓ the triangle stops short, at eye level — (D) The triangle only reaches from the observer’s eye level up to the object; the observer’s own height is the extra distance down to the ground that the triangle skips.
A person $1.5 m$ tall stands on a $28.5 m$ tower and reads the angle of elevation to a flag on a nearby pole as $45^\circ$, at a horizontal distance of $20 m$. Solving $\tan 45^\circ = h/20$ gives $h = 20$ m. A student reports the flag’s height above the ground as $20$ m. What is missing?
nothing — $20$ m is already the full height above the ground
only the tower’s $28.5$ m needs adding, giving $48.5$ m
only the person’s $1.5$ m needs adding, giving $21.5$ m
the eye’s own height above the ground, $30$ m
Check your answer
nothing — $20$ m is already the full height above the ground — $20$ m is only the rise above the observer’s eye; the eye itself sits $30$ m up, at $28.5 + 1.5$ m, and that must be added
only the tower’s $28.5$ m needs adding, giving $48.5$ m — the eye level sits above both the tower and the person standing on it — $28.5 + 1.5 = 30$ m — not the tower alone
only the person’s $1.5$ m needs adding, giving $21.5$ m — the tower itself, not just the person on top of it, sits below the observer’s eye; both heights belong in the offset
✓ the eye’s own height above the ground, $30$ m — (D) The observer’s eye sits $28.5 + 1.5 = 30$ m above the ground; that height still has to be added to the $20$ m the triangle gave.
A drone operator’s eyes sit $48$ m above sea level (the cliff’s height and the operator’s own height already added together). Looking down at a boat, the angle of depression is $30^\circ$. Which distance does $\tan 30^\circ = 48/d$ solve for?
the operator’s own height above the cliff
the horizontal distance to the boat
the length of the line of sight itself
the cliff’s height alone, without the operator
Check your answer
the operator’s own height above the cliff — the operator’s own height is already included in the $48$ m; the unknown $d$ here is the horizontal distance to the boat
✓ the horizontal distance to the boat — (B) $d$ is the horizontal distance to the boat, along the sea, once the operator’s own eye height is already folded into the $48$ m.
the length of the line of sight itself — the line of sight is the hypotenuse of this triangle; $d$ here is the horizontal leg, a different length
the cliff’s height alone, without the operator — the cliff height and the operator’s own height are already combined into the $48$ m eye level; $d$ is the separate horizontal unknown
Once we draw and label the right triangle, one question decides the ratio to use. Which two sides does the problem actually connect, relative to the marked angle?
Look at the triangle. If both the known and the wanted side are legs, opposite and adjacent, we use $\tan$ or $\cot$ to connect them. If one of them is the hypotenuse instead, we use $\sin$, $\cos$, or their reciprocals.
If the hypotenuse is unknown and one leg is known, use $\sin$ or $\cos$. $\tan$ never involves the hypotenuse, so it can never be the answer there.
Your turn: a slide is $5$ m long and meets the ground at $30^\circ$. Find the horizontal distance it covers. (Answer: $4.33$ m, since the length is the hypotenuse and the ground distance is adjacent, so $\cos 30^\circ = d/5$.)
The same triangle appears twice: once with its two legs highlighted, once with a leg and the hypotenuse, showing which pair sets the ratio.
Check yourself
If both the known side and the wanted side are legs of the right triangle (not the hypotenuse), which ratio connects them?
$\sin$ or $\cos$
only $\sin$
any of the six ratios equally
$\tan$ or $\cot$
Check your answer
$\sin$ or $\cos$ — sine and cosine both need the hypotenuse in the ratio; two legs on their own call for tangent or cotangent instead
only $\sin$ — sine needs the hypotenuse in its ratio, which is not among the two legs given here
any of the six ratios equally — the sides given fix which ratio applies — two legs call specifically for tangent or cotangent, not any ratio at random
✓ $\tan$ or $\cot$ — (D) $\tan$ and $\cot$ are the only ratios that connect two legs without involving the hypotenuse.
Why does a problem giving the hypotenuse and one leg call for $\sin$ or $\cos$, rather than $\tan$?
$\tan$ only works for angles under $45^\circ$
$\tan$ gives a less accurate answer than $\sin$ or $\cos$
the hypotenuse can always be swapped for a leg
$\tan$ never involves the hypotenuse
Check your answer
$\tan$ only works for angles under $45^\circ$ — tangent applies to any acute angle used in this chapter; there is no restriction to angles under $45^\circ$
$\tan$ gives a less accurate answer than $\sin$ or $\cos$ — no ratio here is more or less accurate than another; the choice is fixed by which sides the problem hands you
the hypotenuse can always be swapped for a leg — the hypotenuse is a specific side, always the longest, and cannot stand in for one of the two legs
✓ $\tan$ never involves the hypotenuse — (D) Tangent’s ratio is opposite over adjacent, two legs, so it never involves the hypotenuse a problem like this one hands you.
A pole’s height is $10 m$, and the angle of elevation to its top from a point on the ground is $30^\circ$, with the height and the ground distance being the two legs. A student writes $\sin 30^\circ = 10/d$ and gets $d = 20$ m. What went wrong?
nothing — $d = 20$ m is correct
the angle should have been $60^\circ$, not $30^\circ$
the height and distance should be swapped in the ratio
$\tan$, not $\sin$, connects two legs; $d = 10 \sqrt{3}$ m
Check your answer
nothing — $d = 20$ m is correct — sine needs a hypotenuse, which is not one of the two known sides here; using it gives the wrong distance, $20$ m instead of $10 \sqrt{3}$ m
the angle should have been $60^\circ$, not $30^\circ$ — the $30^\circ$ angle is exactly as given; the mistake is picking sine for two legs, not misreading the angle
the height and distance should be swapped in the ratio — swapping which leg sits on top does not fix using sine at all; two legs call for tangent, not any arrangement of sine
✓ $\tan$, not $\sin$, connects two legs; $d = 10 \sqrt{3}$ m — (D) Two legs are given, so tangent, not sine, is the ratio that applies; correctly solving $\tan 30^\circ = 10/d$ gives $d = 10 \sqrt{3}$ m.
A ladder’s foot is $6 m$ from a wall, making $60^\circ$ with the ground. Since $\cot 60^\circ = 6/h$, what is $h$?
the length of the ladder itself
$6 \sqrt{3}$ m, the height up the wall
the distance from the wall, again
the angle the ladder makes with the wall
Check your answer
the length of the ladder itself — the ladder is the hypotenuse; $h$ in this equation is the height up the wall, a different leg
✓ $6 \sqrt{3}$ m, the height up the wall — (B) Solving $\cot 60^\circ = 6/h$ with $\cot 60^\circ = 1/\sqrt{3}$ gives $h = 6 \sqrt{3}$ m, the height up the wall.
the distance from the wall, again — the $6$ m distance from the wall is already given and used in the equation; $h$ is the separate unknown height
the angle the ladder makes with the wall — $h$ is a length, the height up the wall — the equation was never set up to find a second angle
Some problems hand over two angles instead of one. These need two right triangles to answer. The syllabus allows at most two, usually sharing one side: the same base point, or the same vertical line.
We write one trigonometric equation from each triangle. Then we solve the two equations together for the unknown height or distance. Neither equation alone pins the answer down. Only both together do.
A third right triangle is outside this syllabus’s cap. If a problem seems to need one, the set-up has been misread.
Your turn: a $6$ m pole stands between two people on level ground, in line with its foot. One sees its top at $45^\circ$, the other at $60^\circ$. How far apart are they? (Answer: $6 + 2 \sqrt{3}$ m, about $9.46$ m, since the two distances are $6$ m and $6/\sqrt{3}$ m.)
Check yourself
This chapter allows at most two right triangles in one problem, usually sharing
the same angle at every vertex
the same base point or vertical line
the same hypotenuse
no side at all — they are entirely separate
Check your answer
the same angle at every vertex — the two triangles usually differ in their angle at the observer; what they share is a side, not every angle
✓ the same base point or vertical line — (B) The two triangles usually share one side — the same base point or the same vertical line.
the same hypotenuse — the line of sight differs for each sighting, so the hypotenuse is usually not the shared side; the base or vertical is
no side at all — they are entirely separate — a shared side is exactly what lets the two triangles’ equations be solved together; without one, they could not be combined
Why does a two-triangle problem need one trigonometric equation from each triangle, rather than one equation overall?
the problem always has two unknown angles
one triangle’s answer is always wrong
each triangle has its own angle
to check the same answer twice, for accuracy
Check your answer
the problem always has two unknown angles — usually the angles are given and it is two lengths that are unknown — two equations are needed to solve for both
one triangle’s answer is always wrong — neither triangle’s equation is incorrect; each one is correct and contributes a different piece of information
✓ each triangle has its own angle — (C) Each triangle carries its own marked angle, and each angle produces its own equation, so two triangles need two equations.
to check the same answer twice, for accuracy — the two equations are solved together for two different unknowns, not used to double-check a single answer
A problem gives the angle of elevation to the bottom of a flag and, separately, to its top, both from the same ground point. What tells you this needs two triangles rather than one?
the flag has a different colour from the building
only one angle is ever needed for a flagpole problem
two angles, two heights, from one point
the ground point is closer to the flag than usual
Check your answer
the flag has a different colour from the building — colour is not part of the triangle at all; what matters is the two separate angles given, each to a different height
only one angle is ever needed for a flagpole problem — a flagpole problem can need only one angle, but this one hands you two, one to each height — that is the two-triangle signal
✓ two angles, two heights, from one point — (C) Two separate angles are given from the same point, to two different heights — that is exactly the two-triangle signal.
the ground point is closer to the flag than usual — how close the ground point is makes no difference; it is the two separate given angles that call for two triangles
One tower is seen from two ground points, sixty degrees from the near one and thirty from the far, sharing one vertical side.
A SCHOLAR INDIA REMEMBERS
One angle gave you one triangle, with two unknowns in it. What now? Walk closer and measure again. The second angle gives a second triangle. Both triangles share the same height. Each triangle gives one equation. Two equations find both unknowns.
A student says: “the angle of elevation is measured between the line of sight and the tower itself, not the horizontal.” For a tower $15 m$ away with an elevation of $60^\circ$, what does this mistake actually give?
the same height, $15 \sqrt{3}$ m, since the ratio still works out
an undefined result, since the vertical has no angle to the tower
a taller tower than the true height, since the angle looks bigger
$5 \sqrt{3}$ m, instead of the true $15 \sqrt{3}$ m
Check your answer
the same height, $15 \sqrt{3}$ m, since the ratio still works out — swapping the reference to the vertical secretly changes the angle used in the ratio, from $60^\circ$ to its complement $30^\circ$ — the height comes out different, not the same
an undefined result, since the vertical has no angle to the tower — the mistake does not produce nothing — it silently uses $90^\circ - 60^\circ = 30^\circ$ in place of $60^\circ$, giving a smaller, wrong height
a taller tower than the true height, since the angle looks bigger — the substituted angle, $30^\circ$, is smaller than the true $60^\circ$; the mistaken height comes out shorter, not taller
✓ $5 \sqrt{3}$ m, instead of the true $15 \sqrt{3}$ m — (D) Measuring from the vertical instead of the horizontal silently swaps in the complementary angle, $30^\circ$, giving $15 \cdot \tan 30^\circ = 5 \sqrt{3}$ m in place of the true $15 \cdot \tan 60^\circ = 15 \sqrt{3}$ m.
A tower is $18 m$ from an observer, and the true angle of elevation of its top, measured from the horizontal, is $30^\circ$. A student instead measures from the tower’s own vertical line and plugs the same $30^\circ$ into $\tan$ from that wrong reference. What wrong height results?
$6 \sqrt{3}$ m — the same as the true height
$18$ m exactly, using the distance with no ratio applied
a smaller height than the true one, since $30^\circ$ is shallow
$18 \sqrt{3}$ m, far more than the true $6 \sqrt{3}$ m
Check your answer
$6 \sqrt{3}$ m — the same as the true height — measuring from the vertical instead of the horizontal effectively uses the complementary angle, $60^\circ$, giving $18 \sqrt{3}$ m — not the true $6 \sqrt{3}$ m
$18$ m exactly, using the distance with no ratio applied — the wrong reference does not erase the ratio; it flips which angle is effectively used, giving $18 \sqrt{3}$ m, not the bare distance
a smaller height than the true one, since $30^\circ$ is shallow — the wrong reference secretly swaps in the complementary $60^\circ$, which is larger, not smaller — the mistaken height comes out bigger, at $18 \sqrt{3}$ m
✓ $18 \sqrt{3}$ m, far more than the true $6 \sqrt{3}$ m — (D) Measuring from the vertical instead of the horizontal effectively uses the complementary angle, $60^\circ$, giving $18 \cdot \tan 60^\circ = 18 \sqrt{3}$ m in place of the true $18 \cdot \tan 30^\circ = 6 \sqrt{3}$ m.
From the top of a $20 m$ tower, the true angle of depression of a car, measured from the horizontal, is $30^\circ$, giving a true distance of $20 \sqrt{3}$ m. A student instead treats the angle as measured from the tower’s vertical line. What distance does this mistake give?
$20 \sqrt{3}$ m — the same as the true distance
a larger distance, since the tower is very tall
$20/\sqrt{3}$ m, much less than $20 \sqrt{3}$ m
no valid distance, since depression cannot use a vertical reference
Check your answer
$20 \sqrt{3}$ m — the same as the true distance — measuring from the vertical swaps in the complementary $60^\circ$ in place of $30^\circ$, giving $20/\sqrt{3}$ m, not the true $20 \sqrt{3}$ m
a larger distance, since the tower is very tall — the height stays $20$ m either way; the mistake swaps in the complementary $60^\circ$, which gives a smaller distance, not a larger one
✓ $20/\sqrt{3}$ m, much less than $20 \sqrt{3}$ m — (C) Measuring from the vertical swaps in the complementary $60^\circ$ in place of $30^\circ$, giving $20/\tan 60^\circ = 20/\sqrt{3}$ m, far short of the true $20 \sqrt{3}$ m.
no valid distance, since depression cannot use a vertical reference — the mistake does not block a computation — it quietly uses $60^\circ$ where $30^\circ$ belongs, giving a smaller, wrong distance
For any tower and any true elevation angle $\theta$, the “measured from the vertical” mistake always produces the result that the complement $90^\circ - \theta$ would give. Why does this connect the vertical-reference mistake to picking the wrong alternate angle in a two-point problem?
they are unrelated mistakes with different causes
one mistake changes the distance, the other changes the height
both slips swap the correct angle for its complement
only the vertical-reference mistake actually involves an angle
Check your answer
they are unrelated mistakes with different causes — both mistakes compute the same wrong number, the complementary angle’s result — that is too specific to be an unrelated coincidence
one mistake changes the distance, the other changes the height — both mistakes change the same thing — the angle plugged into the ratio — from $\theta$ to $90^\circ - \theta$; the resulting error just shows up in whichever side is being solved for
✓ both slips swap the correct angle for its complement — (C) Both mistakes swap the correct angle in the triangle for its complement, whether the slip is about which line is the reference or about which angle equals which.
only the vertical-reference mistake actually involves an angle — the alternate-angle slip is also, at bottom, using the complementary angle instead of the correct one — the same underlying swap
A tower stands $10 m$ from an observer whose angle of elevation reading is $45^\circ$. A friend claims: “that angle is measured against the tower’s own vertical side.” Is the friend right?
yes — elevation is always read against whatever is vertical nearby
it depends on which side of the tower the observer stands
no — it is measured from the horizontal
Check your answer
yes — elevation is always read against whatever is vertical nearby — the angle of elevation is defined against the horizontal at eye level, never against a nearby vertical object
it depends on which side of the tower the observer stands — standing on a different side changes the distance and possibly the angle’s size, but the reference line is always the horizontal, from any side
✓ no — it is measured from the horizontal — (C) The friend is wrong; the angle of elevation is always measured from the horizontal at the observer’s eye, never against the tower’s own vertical side.
A common mistake treats the angle as measured from the vertical pole or tower itself, not the horizontal. It is an easy slip. The pole stands right there in the picture, and it feels like the natural reference.
But the reference line is always the horizontal at the observer’s eye. It is never the vertical structure beside it. The two references differ by a full $90^\circ$. Confusing them does not give a slightly wrong answer. It gives a completely different triangle.
For a tower $15$ m away with elevation $60^\circ$, the correct height is $15 \cdot \sqrt{3}$ m. Mistake the vertical for the reference, and it secretly swaps in $30^\circ$ for $60^\circ$, giving only $5 \cdot \sqrt{3}$ m: a different, wrong height.
Your turn: before you solve any elevation or depression problem, name the reference line out loud. (Answer: the horizontal at the observer’s eye, never the pole, tower or cliff face beside it.)
Only the horizontal angle is real; the crossed-out tower face gives the wrong height, five root three metres instead of fifteen.
First try
Tower $15$ m away, angle of elevation $60^\circ$. I put the $60^\circ$ at the top, against the tower, and got $15 \cdot \tan 30^\circ = 5 \sqrt{3}$ m.
Second look
An angle of elevation opens from the horizontal at my eye, never from the tower. So the $60^\circ$ sits at my eye, and the height is $15 \cdot \tan 60^\circ = 15 \sqrt{3}$ m.
Find the horizontal first. Every angle of elevation opens from it.
a boat’s distance from a cliff-top depression
Weaker. A boat is seen from the top of a $60$ m cliff at an angle of depression of $30^\circ$. Measure that angle from the cliff face instead of the horizontal, and the cliff’s height sits adjacent to it: $60 \cdot \tan 30^\circ = 20 \cdot \sqrt{3}$ m. But a $30^\circ$ depression is a shallow look down. The boat should sit farther out than the cliff is tall, not closer. An answer shorter than the cliff’s own height means the angle was taken from the wrong line.
Stronger. The angle of depression, $30^\circ$, is always measured from the horizontal at the observer’s eye, never from the cliff face. In the true triangle, the cliff’s $60$ m height is opposite that angle, and the distance is adjacent. $\tan 30^\circ = 60/x$, so $x = 60 \cdot \sqrt{3}$ m. A shallow $30^\circ$ look down does put the boat well beyond the cliff’s own height, so the size now makes sense.
A tower stands $12 m$ from an observer, with an elevation angle of $60^\circ$ to its top. Which single equation gives the tower’s height $h$?
$\sin 60^\circ = h/12$
$\tan 60^\circ = 12/h$
$\cos 60^\circ = 12/h$
$\tan 60^\circ = h/12$
Check your answer
$\sin 60^\circ = h/12$ — sine pairs the opposite side with the hypotenuse; here both $h$ and $12$ m are legs, which calls for tangent instead
$\tan 60^\circ = 12/h$ — the height $h$ is opposite the marked angle and the $12$ m distance is adjacent to it; $\tan$ takes opposite over adjacent, in that order
$\cos 60^\circ = 12/h$ — cosine needs the hypotenuse in its ratio; with only two legs given, tangent is the ratio that applies
✓ $\tan 60^\circ = h/12$ — (D) $h$ is opposite the $60^\circ$ angle and $12$ m is adjacent to it, so tangent, opposite over adjacent, gives $\tan 60^\circ = h/12$.
Using the same method, a flagpole stands $9 m$ from an observer whose angle of elevation to its top is $60^\circ$. What is the flagpole’s height?
$9/\sqrt{3}$ m
$9$ m, the same as the given distance
$18$ m, double the given distance
$9 \sqrt{3}$ m
Check your answer
$9/\sqrt{3}$ m — $\tan 60^\circ = \sqrt{3}$, not $1/\sqrt{3}$ — inverting the ratio gives a shorter, wrong height
$9$ m, the same as the given distance — height equals distance only when the angle is exactly $45^\circ$; at $60^\circ$, $\tan 60^\circ = \sqrt{3} \neq 1$
$18$ m, double the given distance — $\sqrt{3} \approx 1.73$, not exactly $2$ — the height is $9 \sqrt{3}$ m, not a clean double of the distance
✓ $9 \sqrt{3}$ m — (D) $\tan 60^\circ = h/9$ with $\tan 60^\circ = \sqrt{3}$ gives $h = 9 \sqrt{3}$ m.
After computing a tower’s height as $12 \sqrt{3}$ m from a $12 m$ distance and a $60^\circ$ elevation, a check re-solves $\tan \theta = h/d$ for $\theta$ and should return $60^\circ$. If the check had instead returned $45^\circ$, what would that mean?
the original height, $12 \sqrt{3}$ m, would still be correct
an arithmetic slip was made in $h$ or $d$
the tower would need to be measured again from scratch
$45^\circ$ and $60^\circ$ would both be valid for this triangle
Check your answer
the original height, $12 \sqrt{3}$ m, would still be correct — a check that returns a different angle than the one given means a mistake was made in $h$ or $d$ — the height cannot simply stand as correct
✓ an arithmetic slip was made in $h$ or $d$ — (B) One right triangle has exactly one angle at the observer; a check returning a different angle means an arithmetic slip was made in $h$ or $d$.
the tower would need to be measured again from scratch — the mismatch is a computation issue, found by re-checking the working — it does not mean the real tower needs measuring again
$45^\circ$ and $60^\circ$ would both be valid for this triangle — a single triangle has one fixed angle at the observer; getting a different angle back means the numbers plugged in do not match the setup
Worked example
A tower’s height from one triangle
a tower on level ground has its top at an angle of elevation of $60^\circ$, seen from a point $12$ m from its foot the given information, one known distance and one known angle
the $12$ m distance is adjacent to the marked angle, and the tower’s height is opposite it both are legs of the right triangle, so $\tan$ connects them
$\tan 60^\circ = h/12$, so the height is $12 \cdot \tan 60^\circ = 12 \cdot \sqrt{3}$ m $\tan 60^\circ = \sqrt{3}$, from the standard-angle table in Chapter 8
$(12 \cdot \sqrt{3})/12 = \sqrt{3}$, exactly the value the tangent table gives for $60^\circ$ Check the height by dividing it back by the known base. The $12$s cancel and leave $\sqrt{3}$, exactly the tangent table’s value for $60^\circ$.
line of sight
level with the eye
angle of elevation
height above the eye
Siddharth’s eye, the tank’s top and a point on the tower level with his eye form one right triangle.
Twelve metres and sixty degrees are both parts of one triangle, and the tangent of sixty degrees turns one into the other.
a height from one distance and one angle
Draw the triangle Draw a right triangle. The height is the vertical side, the known distance is the horizontal side, and the right angle sits at the foot.
Mark the angle Mark the given angle at the observer’s point. For a tower $12$ m away, that angle is $60^\circ$.
Name the sides Both the known distance and the wanted height are legs of the triangle, not the hypotenuse.
Pick the ratio Two legs need $\tan$. $\tan 60^\circ = h/12$, so $h = 12 \cdot \sqrt{3}$ m.
Check the size An angle over $45^\circ$ means the height is more than the distance. $12 \cdot \sqrt{3}$ is about $20.8$ m, more than $12$ m, so the size is right.
A rope ties the top of an $8 m$ pole to a ground point, making $30^\circ$ with the ground. Which ratio finds the rope’s length $l$?
$\sin 30^\circ = 8/l$
$\tan 30^\circ = 8/l$
$\cos 30^\circ = 8/l$
$\sin 30^\circ = l/8$
Check your answer
✓ $\sin 30^\circ = 8/l$ — (A) The $8 m$ pole is opposite the $30^\circ$ angle and the rope $l$ is the hypotenuse, so sine, opposite over hypotenuse, gives $\sin 30^\circ = 8/l$.
$\tan 30^\circ = 8/l$ — the rope is the line of sight, the hypotenuse of this triangle — tangent never involves the hypotenuse
$\cos 30^\circ = 8/l$ — the $8 m$ pole is opposite the $30^\circ$ angle at the ground, so sine, not cosine, is the ratio that fits
$\sin 30^\circ = l/8$ — sine is opposite over hypotenuse — the pole’s $8 m$ over the rope’s unknown length $l$, not the other way round
Using the same method, a rope ties the top of a $10 m$ pole to the ground, making $30^\circ$ with the ground. What is the rope’s length?
$5$ m
$10 \sqrt{3}$ m
$10$ m, the same as the pole
$20$ m
Check your answer
$5$ m — $\sin 30^\circ = 1/2$, so the rope is the pole’s height divided by $1/2$ — which multiplies it, giving $20$ m, not $5$ m
$10 \sqrt{3}$ m — $10 \sqrt{3}$ m comes from a tangent-style ratio; this triangle’s rope is the hypotenuse, which needs sine
$10$ m, the same as the pole — a rope slanted at $30^\circ$ is always longer than the vertical pole it is tied to; equal lengths would need the rope to stand straight up
✓ $20$ m — (D) $\sin 30^\circ = 10/l$ with $\sin 30^\circ = 1/2$ gives $l = 20$ m.
A student solving the $8 m$ pole and $30^\circ$ rope problem writes $\sin 30^\circ = l/8$ and gets $l = 4$ m. What is wrong?
the ratio is inverted; $l = 16$ m
nothing — a $4$ m rope for an $8$ m pole is fine
the angle should have been $60^\circ$, not $30^\circ$
$8 m$ should have been used as the rope’s own length
Check your answer
✓ the ratio is inverted; $l = 16$ m — (A) The ratio is written upside down; correctly, $\sin 30^\circ = 8/l$ gives $l = 16$ m.
nothing — a $4$ m rope for an $8$ m pole is fine — a slanted rope is always longer than the vertical pole it hangs from; a $4$ m rope for an $8$ m pole cannot be right
the angle should have been $60^\circ$, not $30^\circ$ — the $30^\circ$ angle is exactly as given; the mistake is writing the ratio upside down, not misreading the angle
$8 m$ should have been used as the rope’s own length — the $8 m$ is the pole’s height, the side opposite the angle — it never stands in for the rope’s own length
Worked example
A rope’s length using sine
a rope ties the top of a pole $8$ m tall to a point on the ground, making an angle of $30^\circ$ with the ground the given information. Let the rope’s length be $l$
the pole’s $8$ m height is opposite the marked angle, and the rope is the hypotenuse the rope runs from the ground to the pole’s top, so it is the longest side, the hypotenuse
$\sin 30^\circ = 8/l$, and $\sin 30^\circ = 1/2$, so $l = 16$ m opposite over hypotenuse is $\sin$. Solving $1/2 = 8/l$ for $l$ gives $16$
$8/16 = 1/2$, exactly $\sin 30^\circ$ again Check the length by substituting it back into the ratio. Eight over sixteen simplifies to one half, the same value $\sin 30^\circ$ has.
The rope is the hypotenuse here, so sine, not tangent, connects the eight metre pole to the rope’s sixteen metre length.
A person $1.6 m$ tall stands on an $18.4 m$ tower, eyes at $20 m$. Looking at a car with a $45^\circ$ angle of depression, which equation finds the horizontal distance $d$ to the car?
$\tan 45^\circ = d/20$
$\tan 45^\circ = 18.4/d$
$\sin 45^\circ = 20/d$
$\tan 45^\circ = 20/d$
Check your answer
$\tan 45^\circ = d/20$ — the $20$ m eye height is opposite the depression angle at the top, and $d$ is adjacent to it — $\tan$ takes opposite over adjacent
$\tan 45^\circ = 18.4/d$ — the full eye height is $18.4 + 1.6 = 20$ m; using the tower alone leaves out the observer’s own height
$\sin 45^\circ = 20/d$ — $d$ is the horizontal leg, not the hypotenuse; tangent connects the two legs directly
✓ $\tan 45^\circ = 20/d$ — (D) The $20 m$ eye height is opposite the depression angle and $d$ is adjacent to it, so $\tan 45^\circ = 20/d$.
Using the same setup — eyes at $20 m$ — but with the angle of depression now $30^\circ$, what is the distance to the car?
$20/\sqrt{3}$ m
$20$ m, the same as at $45^\circ$
$20 \sqrt{3}$ m
$10 \sqrt{3}$ m
Check your answer
$20/\sqrt{3}$ m — $\tan 30^\circ = 1/\sqrt{3}$, so $d = 20/(1/\sqrt{3}) = 20 \sqrt{3}$ m, not $20/\sqrt{3}$ m
$20$ m, the same as at $45^\circ$ — a shallower depression angle over the same eye height means the car is farther away, not the same distance as at $45^\circ$
✓ $20 \sqrt{3}$ m — (C) $\tan 30^\circ = 20/d$ with $\tan 30^\circ = 1/\sqrt{3}$ gives $d = 20 \sqrt{3}$ m.
$10 \sqrt{3}$ m — the full $20$ m eye height belongs in the ratio; there is no reason to halve it first
For the same $1.6 m$ person on the $18.4 m$ tower, a student solves $\tan 30^\circ = h/d$ using only the tower’s $18.4 m$ for $h$, ignoring the person’s height, and reports the distance as $18.4 \sqrt{3}$ m. What is wrong?
nothing — the tower’s height is all that matters here
the angle should be applied to $1.6 m$ instead of $18.4 m$
the true eye height is $20$ m, not $18.4$ m
the distance cannot be found without the car’s own height
Check your answer
nothing — the tower’s height is all that matters here — the observer’s eye sits $1.6$ m above the tower’s own top; the full $20$ m eye height belongs in the ratio, not $18.4$ m alone
the angle should be applied to $1.6 m$ instead of $18.4 m$ — neither height stands alone; the tower and the person’s own height add together to $20$ m before the ratio is applied
✓ the true eye height is $20$ m, not $18.4$ m — (C) The observer’s eye sits $1.6$ m above the tower’s own top; the full $20$ m eye height belongs in the ratio, giving $20 \sqrt{3}$ m.
the distance cannot be found without the car’s own height — the car sits at ground level, contributing no extra height; only the observer’s own eye height, $20$ m, is needed
The tower-and-person eye-level setup and an earlier elevation example at ground level (no offset) both end with the same equation shape, $\tan \theta = \text{opposite}/\text{adjacent}$. What is the one real difference between them?
the ratio used is completely different in each case
one uses elevation and the other cannot use depression
whether the observer’s own height must be added first
the two examples cannot be compared at all
Check your answer
the ratio used is completely different in each case — both examples end up using $\tan$; the difference is not the ratio but whether an eye-level offset needs adding first
one uses elevation and the other cannot use depression — the offset rule is not tied to elevation or depression specifically — either angle type can involve an observer standing above the ground
✓ whether the observer’s own height must be added first — (C) The two examples use the same ratio shape; the real difference is whether an eye-level offset needs adding before or after solving it.
the two examples cannot be compared at all — both are solved by marking one right triangle and choosing a ratio — that shared method is exactly what makes them comparable
Worked example
Depression with the observer’s own height
a person $1.6$ m tall stands on a tower $18.4$ m high, so the eyes sit $20$ m above the ground $18.4 + 1.6 = 20$. We add the observer’s own height to the tower before drawing the triangle
a car is seen at an angle of depression of $45^\circ$ the given angle, measured from the observer’s horizontal at eye level
the $20$ m eye height is opposite the marked angle at the car’s own elevation angle, which equals $45^\circ$ by alternate angles the angle of depression from the eye equals the angle of elevation from the car
$\tan 45^\circ = 20/d$, and $\tan 45^\circ = 1$, so $d = 20$ m the horizontal distance equals the eye height exactly when the ratio is $1$
$20/20 = 1$, exactly $\tan 45^\circ$ Check the distance by dividing it back by the eye height. Twenty over twenty is one. The car sits exactly as far out as the eyes are high, exactly what a $45^\circ$ angle means.
The tower and the observer’s own height stack to twenty metres of eye level, matching the twenty metre distance to the car.
a distance from an eye height and one angle
Add the heights $18.4 + 1.6 = 20$ m. The tower’s height and the observer’s own height together give the eye’s height above the ground.
Draw from the eye Draw the right triangle from the $20$ m eye level down to the car, not from the tower’s foot.
Mark the angle Mark the given angle of depression, $45^\circ$, at the eye.
Use alternate angles The $45^\circ$ depression at the eye equals the $45^\circ$ elevation at the car.
Pick the ratio and solve $\tan 45^\circ = 1$, so the horizontal distance equals the $20$ m eye height.
A $20 m$ building carries a flagpole on its roof. From point $P$, the building’s top has elevation $30^\circ$ and the flagpole’s top has elevation $45^\circ$. Which equation finds $d$, the distance from $P$ to the building’s foot?
$\tan 45^\circ = 20/d$
$\tan 30^\circ = 20/d$
$\tan 30^\circ = d/20$
$\sin 30^\circ = 20/d$
Check your answer
$\tan 45^\circ = 20/d$ — the building’s own height, $20$ m, pairs with the building’s own angle, $30^\circ$ — not with the flagpole’s $45^\circ$
✓ $\tan 30^\circ = 20/d$ — (B) The building’s own $20$ m pairs with its own $30^\circ$ angle, giving $\tan 30^\circ = 20/d$.
$\tan 30^\circ = d/20$ — the building’s $20$ m is opposite the $30^\circ$ angle at $P$, and $d$ is adjacent to it — tangent takes opposite over adjacent
$\sin 30^\circ = 20/d$ — $d$ is the horizontal leg, not the hypotenuse; the building’s height and the ground distance call for tangent
Using the same method, a $10 m$ building carries a flagpole. From point $P$, the building’s top has elevation $30^\circ$ and the flagpole’s top has elevation $60^\circ$. What is the flagpole’s length?
$10 \sqrt{3} - 10$ m
$10$ m, the same as the building
$30$ m, the total height including the building
$20$ m
Check your answer
$10 \sqrt{3} - 10$ m — that expression matches a $30^\circ$/$45^\circ$ pair from a different example; with $30^\circ$ and $60^\circ$ here, the flagpole works out to a clean $20$ m
$10$ m, the same as the building — the flagpole’s length comes from the two given angles, $30^\circ$ and $60^\circ$ — nothing here makes it equal to the building’s own height
$30$ m, the total height including the building — $30$ m is the total height to the flagpole’s top; the question asks for the flagpole’s own length, which is $30 - 10 = 20$ m
✓ $20$ m — (D) $d = 10 \sqrt{3}$ m from the first triangle; then $\tan 60^\circ = (10+x)/d$ gives $10+x = 30$, so the flagpole is $x = 20$ m.
Solving the $20 m$ building and flagpole problem, a student finds $d = 20 \sqrt{3}$ m from the $30^\circ$ triangle, then writes $\tan 45^\circ = x/(20 \sqrt{3})$ using $x$ alone (the flagpole) instead of the total height, and reports $x = 20 \sqrt{3}$ m. What is wrong?
nothing — $x = 20 \sqrt{3}$ m is the flagpole’s correct length
$\tan 45^\circ$ needs the total height, $20+x$
the distance $d$ should have been recalculated for the second triangle
the $30^\circ$ angle should have been used again instead of $45^\circ$
Check your answer
nothing — $x = 20 \sqrt{3}$ m is the flagpole’s correct length — the $45^\circ$ angle sights the flagpole’s top, so its triangle’s vertical side is the building plus the flagpole together, not the flagpole alone
✓ $\tan 45^\circ$ needs the total height, $20+x$ — (B) The $45^\circ$ angle sights the flagpole’s top, so its equation needs the total height, $20+x$, not the flagpole alone.
the distance $d$ should have been recalculated for the second triangle — both triangles share the same base point $P$ and the same distance $d$; what changes between them is the vertical side, not $d$
the $30^\circ$ angle should have been used again instead of $45^\circ$ — the $45^\circ$ triangle correctly uses $45^\circ$; the mistake is in which height that equation is set equal to
Worked example
A building and its flagpole
a building $20$ m tall carries a flagpole on its roof, seen from a ground point $P$ at elevation $30^\circ$ to the roof and $45^\circ$ to the flag’s top the given information. Let $d$ be the distance from $P$ to the foot of the building, and $x$ the flagpole’s length
in the smaller triangle, $\tan 30^\circ = 20/d$, so $d = 20 \cdot \sqrt{3}$ m the building’s own height and distance form one right triangle, solved first
in the larger triangle, $\tan 45^\circ = (20 + x)/d = 1$, so $20 + x = d$ the flagpole’s top adds $x$ to the building’s height, over the same distance $d$
$x = 20 \cdot \sqrt{3} - 20 = 20 \cdot (\sqrt{3} - 1) \approx 14.6$ m substitute the value of $d$ found from the smaller triangle
$20 + x = 20 + 20 \cdot (\sqrt{3} - 1) = 20 \cdot \sqrt{3} = d$ Check by adding the flagpole’s length back onto the building’s height. The $20$s cancel, leaving $20 \sqrt{3}$, exactly the distance $d$ found from the smaller triangle.
two triangles sharing one side
Draw both triangles Draw one triangle to the building’s roof and a taller one to the flagpole’s top, both from the same ground point $P$.
Mark both angles Mark $30^\circ$ to the roof and $45^\circ$ to the flag, both at $P$.
Name the shared side Both triangles share the same base, the distance $d$ from $P$ to the building’s foot.
Solve the smaller triangle $\tan 30^\circ = 20/d$, so $d = 20 \cdot \sqrt{3}$ m.
Solve the larger triangle $\tan 45^\circ = (20 + x)/d$, using the same $d$. Solve for $x$.
Check the size A bigger angle to the flag means $x$ must be positive. Here $x \approx 14.6$ m, which checks out.
Two angles from one ground point, thirty degrees to the roof and forty-five to the flag, are enough to find the flagpole’s length.
A shorter $8 m$ building stands near a taller one. From the taller building’s top, the depression angle to the shorter one’s top is $30^\circ$, and to its foot is $60^\circ$. By alternate angles, what does the $60^\circ$ depression angle equal, seen from the shorter building’s foot?
an elevation angle of $30^\circ$
an elevation angle of $60^\circ$
the complement of $60^\circ$, which is $30^\circ$
no fixed relationship — it depends on the buildings’ heights
Check your answer
an elevation angle of $30^\circ$ — each depression angle equals its own alternate elevation angle — the $60^\circ$ depression to the foot pairs with a $60^\circ$ elevation from the foot, not $30^\circ$
✓ an elevation angle of $60^\circ$ — (B) The $60^\circ$ depression angle to the foot equals its own alternate elevation angle, also $60^\circ$, seen from the shorter building’s foot.
the complement of $60^\circ$, which is $30^\circ$ — alternate angles are equal, not complementary — the elevation angle from the foot is $60^\circ$, the same as the depression angle, not its complement
no fixed relationship — it depends on the buildings’ heights — the equal-angle relationship comes from the two horizontals being parallel; it holds whatever the two buildings’ heights turn out to be
Using the same method — a shorter $6 m$ building near a taller one, with depression angles of $30^\circ$ to its top and $60^\circ$ to its foot — what is the height of the taller building?
$12$ m
$6$ m, the same as the shorter building
$9$ m
$3 \sqrt{3}$ m
Check your answer
$12$ m — $12$ m was the answer for an $8$ m shorter building; with a $6$ m shorter building, solving the two equations together gives $9$ m
$6$ m, the same as the shorter building — the taller building’s height comes from solving the two triangles together; it is not simply set equal to the shorter one’s height
✓ $9$ m — (C) Solving $\tan 60^\circ = H/x$ and $\tan 30^\circ = (H-6)/x$ together gives $x = 3 \sqrt{3}$ m and $H = 9$ m.
$3 \sqrt{3}$ m — $3 \sqrt{3}$ m is the distance between the two buildings; the question asks for the taller building’s height, which is $9$ m
Solving the $8 m$ shorter building problem, a student writes $\tan 30^\circ = H/x$ for the foot and $\tan 60^\circ = (H-8)/x$ for the top — swapping the two angles from the original setup. What goes wrong?
nothing — the equations can be paired with either angle
the equations should use $8$ instead of $H-8$
$x$ should be different in each equation
the steeper $60^\circ$ belongs to the nearer foot
Check your answer
nothing — the equations can be paired with either angle — each depression angle is fixed to what it was sighted to — $60^\circ$ to the nearer foot, $30^\circ$ to the farther top — they cannot be swapped
the equations should use $8$ instead of $H-8$ — the top equation needs the remaining vertical drop, $H-8$, above the shorter building; the fix here is the swapped angles, not the $H-8$ term
$x$ should be different in each equation — both triangles correctly share the same distance $x$; the actual mistake is the two depression angles being swapped between them
✓ the steeper $60^\circ$ belongs to the nearer foot — (D) The steeper angle, $60^\circ$, was sighted to the nearer foot, not the farther top; the two angles are swapped between the equations.
The two-buildings problem and the building-and-flagpole problem both solve two equations together. What is different about which unknown the alternate-angle equality supplies?
it converts depression to elevation angles
the flagpole problem needs alternate angles in the same way
neither problem actually needs two equations
the flagpole problem needs alternate angles instead
Check your answer
✓ it converts depression to elevation angles — (A) The two-buildings problem needs the alternate-angle equality to convert its depression angles to elevation angles before the two equations are written.
the flagpole problem needs alternate angles in the same way — the flagpole problem’s two elevation angles are already given from the same point $P$; there is no depression angle to convert there
neither problem actually needs two equations — both problems solve two trigonometric equations together — one for each triangle — to pin down two unknowns
the flagpole problem needs alternate angles instead — it is the two-buildings problem that converts a depression angle to its alternate elevation angle; the flagpole problem never needs that step
Worked example
Two buildings, by angle of depression
a shorter building $8$ m tall stands near a taller one, depression $30^\circ$ to its top and $60^\circ$ to its foot, both measured from the taller building’s top the given information. Let $x$ be the distance between the buildings, and $H$ the taller building’s height
by alternate angles, these depression angles equal the elevation angles measured from the shorter building’s top and foot the alternate-angle equality proved earlier, applied twice
from the foot: $\tan 60^\circ = H/x$ the full height $H$ and the shared distance $x$ form one right triangle
from the top: the remaining vertical drop is $H - 8$, so $\tan 30^\circ = (H - 8)/x$ the shorter building’s own height, $8$ m, is subtracted from $H$ first
solving the two equations together gives $x = 4 \cdot \sqrt{3}$ m and $H = 12$ m substitute $H = x \cdot \sqrt{3}$ from the first equation into the second, then solve for $x$
$H/x = 12/(4 \cdot \sqrt{3}) = \sqrt{3}$, matching $\tan 60^\circ$, and $(H - 8)/x = 4/(4 \cdot \sqrt{3}) = 1/\sqrt{3}$, matching $\tan 30^\circ$ *Check both original equations by substituting $x = 4 \sqrt{3}$ and $H = 12$ back in.* Both ratios land exactly on the tangents the problem started from.
The depression angles from the taller roof match the elevation angles at the shorter building, which is how both triangles get solved.
This chapter’s method, every time, is: draw and label one right triangle, mark the given angle, and then
measure the object directly, to double-check
add a second, unrelated triangle for safety
choose the connecting ratio
guess the answer and adjust it afterward
Check your answer
measure the object directly, to double-check — the method exists precisely because the object cannot be measured directly; the ratio, not a direct check, gives the answer
add a second, unrelated triangle for safety — a second triangle is added only when a problem needs it, and shares a side with the first — never as a routine extra step
✓ choose the connecting ratio — (C) The third step is choosing the ratio that connects the known side to the wanted one.
guess the answer and adjust it afterward — the ratio gives the answer directly from the known side and angle; nothing here is guessed and adjusted afterward
This chapter never introduces a new trigonometric ratio or identity. What does it add to what Chapter 8 already built?
a fourth trigonometric ratio, for slanted lines specifically
a rule for when $\sin \theta$ and $\cos \theta$ are equal
a correction to how elevation and depression are defined
a method for aiming those ratios at real problems
Check your answer
a fourth trigonometric ratio, for slanted lines specifically — no new ratio is introduced here; the same six ratios from Chapter 8 are reused throughout this chapter
a rule for when $\sin \theta$ and $\cos \theta$ are equal — any such identity belongs among Chapter 8’s own trigonometric results; this chapter only applies ratios, it does not prove new ones
a correction to how elevation and depression are defined — elevation and depression keep the same definition throughout; this chapter never revises them
✓ a method for aiming those ratios at real problems — (D) This chapter adds a method for aiming Chapter 8’s own ratios at real measurement problems, without adding a new ratio or identity.
Why are the same three steps — triangle, angle, ratio — enough for every problem in this chapter, from a single tower to two buildings at once?
a two-triangle problem repeats the same method, sharing a side
harder problems secretly need a fourth trigonometric ratio
the observer’s height is ignored once a second triangle is added
two-triangle problems use a completely different method
Check your answer
✓ a two-triangle problem repeats the same method, sharing a side — (A) A two-triangle problem is the same triangle-angle-ratio method run twice, on two triangles that share one side.
harder problems secretly need a fourth trigonometric ratio — every problem here, however many triangles it needs, uses only the same ratios already built in Chapter 8 — nothing extra is required
the observer’s height is ignored once a second triangle is added — the eye-level offset applies whenever the observer is raised above the ground, regardless of whether one triangle or two are needed
two-triangle problems use a completely different method — a two-triangle problem runs the very same method twice — draw, mark, choose the ratio — once for each triangle, then combines the results
Check yourself: the whole chapter
A student wants to find the height of a mobile tower without climbing it. Which two measurements does the method in this chapter actually need?
the tower’s shadow length and the time of day
two different distances measured from the tower’s foot, with no angle taken at all
one angle of elevation to the top of the tower, and the distance from the point of measurement to the tower’s foot
two different angles of elevation taken from the same spot, with no distance measured at all, relying on the two angles alone to fix the height
Check your answer
the tower’s shadow length and the time of day — The shadow-and-time method is a different technique altogether. This chapter’s method needs one measured angle and one measured distance, not a shadow length.
two different distances measured from the tower’s foot, with no angle taken at all — Two distances alone give no angle to work with, and the method turns on exactly one measured angle — without it, no triangle can be solved.
✓ one angle of elevation to the top of the tower, and the distance from the point of measurement to the tower’s foot — (C) The method needs exactly one angle — here, the angle of elevation to the tower’s top — and one accessible distance. The right triangle they form gives the unknown height.
two different angles of elevation taken from the same spot, with no distance measured at all, relying on the two angles alone to fix the height — An angle by itself has no length attached to it. Without one measured distance, two angles alone cannot fix an actual height.
An observer, from one spot, looks up at a plane and down at a boat on a lake. What line is each of the two angles measured from?
the angle to the plane is measured from the horizontal. The angle to the boat is measured from the vertical drop down to the lake below
each angle is measured from the observer’s line of sight to the OTHER object. Neither is based on any horizontal line
the plane and the boat each need their own separate horizontal line. They point in different directions
both angles are measured from the same horizontal line at the observer’s eye. The plane’s angle is read upward, the boat’s downward
Check your answer
the angle to the plane is measured from the horizontal. The angle to the boat is measured from the vertical drop down to the lake below — Elevation and depression are both measured from the horizontal at eye level, never from a vertical drop. Bringing in the vertical for the boat’s angle breaks that rule.
each angle is measured from the observer’s line of sight to the OTHER object. Neither is based on any horizontal line — Neither angle is defined by the other’s line of sight. Both are measured against one fixed reference, the horizontal — the plane and the boat never enter each other’s definition.
the plane and the boat each need their own separate horizontal line. They point in different directions — One horizontal, at the observer’s own eye, serves both angles. Looking in different directions does not create a second horizontal.
✓ both angles are measured from the same horizontal line at the observer’s eye. The plane’s angle is read upward, the boat’s downward — (D) Elevation and depression share one reference line, the horizontal at the observer’s eye — the plane’s angle is read upward from it, the boat’s downward.
Every problem in this chapter starts with one right triangle — a vertical side, a horizontal side, and the line of sight as the third side. Where does the right angle sit?
where the line of sight meets the horizontal side
where the vertical side meets the horizontal side
where the line of sight meets the vertical side
at the top of the vertical side, next to the object being viewed
Check your answer
where the line of sight meets the horizontal side — The line of sight is the triangle’s slanted side, the hypotenuse — it meets the horizontal at the marked angle, not at a right angle.
✓ where the vertical side meets the horizontal side — (B) The right angle sits where the vertical side meets the horizontal side — the line of sight is the triangle’s slanted third side, and never forms the right angle itself.
where the line of sight meets the vertical side — The line of sight also meets the vertical side at an angle, not a right angle — the right angle belongs where the two straight sides meet each other.
at the top of the vertical side, next to the object being viewed — The vertical and horizontal sides meet at the base of the object, not at its top — that base is where the right angle goes.
In a right triangle, the side you are given is the hypotenuse, and the side you want is one of the two legs. Which ratio should you use, relative to the marked angle?
$\tan$, since it always connects two sides of the triangle
$\cot$, since it connects the two legs
$\sin$ or $\cos$, since one of the two sides involved is the hypotenuse
any of the three ratios works equally well here, since all of them connect two sides
Check your answer
$\tan$, since it always connects two sides of the triangle — $\tan$ connects the two legs of a right triangle, opposite and adjacent — it never has the hypotenuse in it, so it cannot be the ratio here.
$\cot$, since it connects the two legs — $\cot$ is $\tan$’s reciprocal and shares the same limitation — both legs, never the hypotenuse.
✓ $\sin$ or $\cos$, since one of the two sides involved is the hypotenuse — (C) With the hypotenuse as one of the two sides involved, only $\sin$ or $\cos$ connects it to a leg — $\tan$ and $\cot$ never touch the hypotenuse at all.
any of the three ratios works equally well here, since all of them connect two sides — The ratio is not a free choice. Whenever the hypotenuse is one of the two sides in play, only $\sin$, $\cos$, or their reciprocals apply — never $\tan$ or $\cot$.
A problem gives two different angles of elevation to two different points on the same tower, both measured from the same spot on the ground. How does the standard method in this chapter handle this?
one triangle, combining both angles into a single trigonometric equation
three triangles, one for each possible pairing of the two angles with the shared ground point
two triangles, but a trigonometric equation is written for only one of them
two triangles sharing one side, each giving its own trigonometric equation, solved together
Check your answer
one triangle, combining both angles into a single trigonometric equation — Two different angles to two different points make two separate triangles, each needing its own equation — one combined equation loses information the problem actually gives.
three triangles, one for each possible pairing of the two angles with the shared ground point — This chapter’s problems never need a third right triangle; if a set-up seems to call for one, it has been misread. Two angles from one spot make exactly two triangles.
two triangles, but a trigonometric equation is written for only one of them — Both triangles contribute an equation. Stopping after the first throws away the second angle’s information, leaving the problem half-solved.
✓ two triangles sharing one side, each giving its own trigonometric equation, solved together — (D) Two angles to two points from the same spot give two right triangles sharing one side — write one trigonometric equation from each, then solve the two together.
TRIANGLE: vertical side, horizontal side and line of sight, angle marked at the observer. The tower problem marked its $60^\circ$ angle there, $12$ m from the foot.
RATIO: two legs need $\tan$ or $\cot$; a leg with the hypotenuse needs $\sin$ or $\cos$. The rope problem used $\sin 30^\circ = 8/l$ for its hypotenuse.
EYE LEVEL: add the observer’s own height back after solving the triangle. The tower’s $18.4$ m plus the person’s $1.6$ m made $20$ m of eye height.
ALTERNATE ANGLES: depression from a high point equals elevation from the low point below it. A $30^\circ$ depression from a cliff top gave a $30^\circ$ elevation from the boat.
TWO TRIANGLES: at most two triangles, sharing one side, solved as two equations together. The flagpole came out $20 \cdot (\sqrt{3} - 1) \approx 14.6$ m from both.
Step back. We have run the same three steps every time. Draw and label one right triangle, at most two if the problem needs a second. Mark the given angle of elevation or depression at the observer. Then choose the ratio that connects the known side to the wanted one.
A tower, a rope, a flagpole and two buildings: different pictures, the same triangle underneath each one.
Your turn: an observer $1.2$ m tall stands $5$ m from a pole and measures $60^\circ$ of elevation to its top. Find the pole’s full height. (Answer: $9.86$ m, since $\tan 60^\circ$ gives a rise of $5 \cdot \sqrt{3} \approx 8.66$ m, plus the observer’s $1.2$ m.)
You judge a height or a distance you cannot reach more often than you think. Here are seven places one angle and one triangle give you that number.
Measuring a temple gopuram’s height. You stand $30$ m from the foot of a temple gopuram. Its top has an angle of elevation of $30^\circ$. The maths: the height comes from the horizontal distance and the tangent of the elevation angle. Since $\tan 30^\circ = 1/\sqrt{3}$, the height is $30/\sqrt{3} = 10 \sqrt{3} \approx 17.3$ m.
One angle and one distance, and the tangent turns them into the height you could not climb up to measure.
Estimating a kite’s height from the string. A kite string is let out to $50$ m, taut, at an angle of elevation of $60^\circ$ from the flyer’s hand. The maths: the kite’s height uses the sine ratio, since the taut string is the hypotenuse. Since $\sin 60^\circ = \sqrt{3}/2$, the height is $50 \cdot \sqrt{3}/2 = 25 \sqrt{3} \approx 43.3$ m.
Judging a parked car’s distance from your terrace. You stand on a terrace $12$ m high. You see a car parked on the road at an angle of depression of $30^\circ$. The maths: the angle of depression from the terrace equals the angle of elevation measured from the car looking up. Since $\tan 30^\circ = 1/\sqrt{3}$, the car’s distance is $12 \sqrt{3} \approx 20.8$ m.
Measuring a tree’s height for a school project. A student’s eyes are $1.5$ m above the ground. From $20$ m away, the tree’s top has an elevation of $30^\circ$. The maths: the tree’s full height adds the observer’s own eye level to the triangle’s vertical side. Since $\tan 30^\circ = 1/\sqrt{3}$, the triangle gives $20/\sqrt{3} \approx 11.5$ m. Adding the eyes’ $1.5$ m, the tree is about $13.0$ m tall.
Finding a statue’s pedestal height. A $1.6$ m statue stands on a pedestal in a town square. From one point on the ground, the pedestal’s top is at $45^\circ$ and the statue’s top at $60^\circ$. The maths: two triangles share the same base distance, and each gives one equation for it. With the pedestal $h$ m tall, the base distance is $h$. So $h \sqrt{3} = h + 1.6$, and $h = 1.6/(\sqrt{3} - 1) \approx 2.19$ m.
Estimating a Ferris wheel’s height at a fair. You walk back from a Ferris wheel until its top is at an angle of elevation of $45^\circ$. You are then $40$ m from its base. The maths: a $45^\circ$ elevation makes the height equal the horizontal distance, since $\tan 45^\circ = 1$. So the wheel is $40$ m tall, read off your own paces with no table needed.
Checking a builder’s claimed apartment height. A builder claims an apartment tower is $52$ m tall. From $30$ m away, you measure the top’s angle of elevation as $60^\circ$. The maths: the tangent ratio turns your distance and angle into the tower’s real height, which you can check against the claim. Since $\tan 60^\circ = \sqrt{3}$, the measured height is $30 \sqrt{3} \approx 51.96$ m, close to the claimed $52$ m.
Your turn. You stand $15$ m from the foot of your school’s flagpole. Its top has an angle of elevation of $30^\circ$. How tall is the pole? Answer: Since $\tan 30^\circ = 1/\sqrt{3}$, the height is $15/\sqrt{3} = 5 \sqrt{3} \approx 8.66$ m.
practice A tower stands on level ground. From a point $18$ m from its foot, the angle of elevation of the top of the tower is $60^\circ$. Find the height of the tower. (Worked in full below — read it, then do the next three the same way.)
practice A ladder $12$ m long leans against a vertical wall and makes an angle of $30^\circ$ with the WALL. Find how high up the wall the ladder reaches. (Mark the angle first — this $30^\circ$ is at the top, between the ladder and the wall, so the wall is the side next to it and the ladder is the hypotenuse. That makes the ratio cosine, not tangent.)
practice A girl $1.5$ m tall stands $12$ m from the foot of a lamp post. From her eyes, the angle of elevation of the top of the post is $30^\circ$. Find the height of the post. (Two steps — the triangle gives only the rise ABOVE her eyes, so work that out first and then add her own height.)
practice A statue stands on top of a pedestal $12$ m tall. From a point on the ground, the angle of elevation of the top of the pedestal is $45^\circ$ and of the top of the statue is $60^\circ$. Find the height of the statue. (Two triangles share the same ground distance — use the smaller one to find that distance first, then put it into the larger one.)
practice A ladder leans against a wall so that its foot is $2.5$ m from the wall and it makes an angle of $60^\circ$ with the ground. Find the length of the ladder.
practice A $10$ m ladder rests against a vertical wall, making an angle of $30^\circ$ with the wall itself. Find how far the foot of the ladder is from the wall.
practice A tower stands on level ground. From a point $20$ m from its foot, the angle of elevation of the top is $45^\circ$. Find the height of the tower.
practice A tower stands on level ground. From a point $30$ m from its foot, the angle of elevation of the top is $30^\circ$. Find the height of the tower.
practice A kite’s taut string is $40$ m long and makes an angle of $60^\circ$ with the ground. Find the height of the kite above the ground.
practice Two points lie on the same side of a tower, in line with its foot, $20$ m apart. From the nearer point, the angle of elevation of the tower’s top is $60^\circ$. From the farther point, the angle is $30^\circ$. Find the height of the tower.
practice A building $15$ m tall carries a flagpole on its roof. From a ground point, the angle of elevation of the building’s top is $30^\circ$ and of the flagpole’s top is $45^\circ$. Find the length of the flagpole.
practice A shorter building $10$ m tall stands near a taller one. From the taller building’s top, the angle of depression of the shorter building’s top is $30^\circ$ and of its foot is $60^\circ$. Find the taller building’s height and the distance between them.
practice From the top of a lighthouse $60$ m tall, the angle of depression of a boat is $60^\circ$. Find the boat’s distance from the foot of the lighthouse.
practice From the top of a building $30$ m tall, the angles of depression of two cars on the same road, both on the same side, are $60^\circ$ (nearer car) and $30^\circ$ (farther car). Find the distance between the two cars.
practice An observer $1.6$ m tall stands $30$ m from a building. The angle of elevation of the building’s top, measured from the observer’s eye, is $30^\circ$. Find the building’s height.
practice A person’s eyes are $10$ m above the ground, standing on a tower. The angle of depression of a car on the ground is $30^\circ$. Find the car’s distance from the foot of the tower.
practice A tower stands on one bank of a river. From its top, the angle of depression of a point directly opposite, on the other bank, is $60^\circ$. The angle of depression of a second point, $24$ m further away on the line joining the first point to the foot of the tower, is $30^\circ$. Find the width of the river and the height of the tower.
practice A pole stands on top of a pedestal $5$ m tall. From a ground point, the angle of elevation of the pedestal’s top is $30^\circ$ and of the pole’s top is $60^\circ$. Find the length of the pole.
practice Two poles of equal height stand on either side of a road $80$ m wide. From a point on the road between them, the angles of elevation of the tops are $30^\circ$ and $60^\circ$. Find the height of the poles and the distances of the point from each pole.
Answers
$h = 18 \cdot \sqrt{3}$ m, about $31.2$ m.
$\cos 30^\circ = x/12 = \sqrt{3}/2$, so $x = 6 \cdot \sqrt{3}$ m, about $10.4$ m.
The rise above eye level is $12 \cdot \tan 30^\circ = 12/\sqrt{3} = 4 \cdot \sqrt{3}$ m. Adding her own height, the post is $1.5 + 4 \cdot \sqrt{3}$ m tall, about $8.4$ m.
$\tan 45^\circ = 12/d = 1$, so $d = 12$ m. The full height is $d \cdot \tan 60^\circ = 12 \cdot \sqrt{3}$ m, so the statue is $12 \cdot \sqrt{3} - 12 = 12 \cdot (\sqrt{3} - 1)$ m, about $8.8$ m.
$\cos 60^\circ = 2.5/l$, and $\cos 60^\circ = 1/2$, so $l = 5$ m.
The angle at the top, between the ladder and the wall, is $30^\circ$; the horizontal distance is opposite it. $\sin 30^\circ = d/10 = 1/2$, so $d = 5$ m.
$\tan 45^\circ = h/20 = 1$, so $h = 20$ m.
$\tan 30^\circ = h/30 = 1/\sqrt{3}$, so $h = 30/\sqrt{3} = 10 \cdot \sqrt{3}$ m, about $17.3$ m.
$\sin 60^\circ = h/40 = \sqrt{3}/2$, so $h = 20 \cdot \sqrt{3}$ m, about $34.6$ m.
Let the nearer distance be $a$. $\tan 60^\circ = h/a$ and $\tan 30^\circ = h/(a + 20)$. Solving together gives $a = 10$ m and $h = 10 \cdot \sqrt{3}$ m, about $17.3$ m.
$d = 15/(\tan 30^\circ) = 15 \cdot \sqrt{3}$ m. The flagpole’s own length is $d \cdot \tan 45^\circ - 15 = 15 \cdot \sqrt{3} - 15 = 15 \cdot (\sqrt{3} - 1)$ m, about $11.0$ m.
Let $x$ be the distance and $H$ the taller building’s height. $H = x \cdot \sqrt{3}$ and $H - 10 = x/\sqrt{3}$. Solving together gives $x = 5 \cdot \sqrt{3}$ m, about $8.7$ m, and $H = 15$ m.
$\tan 60^\circ = 60/d = \sqrt{3}$, so $d = 60/\sqrt{3} = 20 \cdot \sqrt{3}$ m, about $34.6$ m.
Nearer car: $\tan 60^\circ = 30/a$, so $a = 10 \cdot \sqrt{3}$ m. Farther car: $\tan 30^\circ = 30/b$, so $b = 30 \cdot \sqrt{3}$ m. The distance between them is $b - a = 20 \cdot \sqrt{3}$ m, about $34.6$ m.
The rise above eye level is $30 \cdot \tan 30^\circ = 10 \cdot \sqrt{3}$ m. Adding the observer’s own height, the building is $10 \cdot \sqrt{3} + 1.6$ m tall, about $18.9$ m.
$\tan 30^\circ = 10/d = 1/\sqrt{3}$, so $d = 10 \cdot \sqrt{3}$ m, about $17.3$ m.
Let the width be $a$. $\tan 60^\circ = h/a$ and $\tan 30^\circ = h/(a + 24)$. Solving together gives $a = 12$ m and $h = 12 \cdot \sqrt{3}$ m, about $20.8$ m.
$d = 5/(\tan 30^\circ) = 5 \cdot \sqrt{3}$ m. The full height is $d \cdot \tan 60^\circ = 5 \cdot \sqrt{3} \cdot \sqrt{3} = 15$ m, so the pole itself is $15 - 5 = 10$ m.
Let the distance from the nearer pole be $a$. $a \cdot \tan 60^\circ = (80 - a) \cdot \tan 30^\circ$ gives $a = 20$ m. The height is $20 \cdot \tan 60^\circ = 20 \cdot \sqrt{3}$ m, about $34.6$ m; the two distances are $20$ m and $60$ m.
Two equal poles stand at each end of an 80 metre road, sighted from one point between them at thirty and sixty degrees.
Exercise 9.1 — further practice
practice In a right triangle formed for a heights-and-distances problem, the hypotenuse is unknown and one leg is known. Which trigonometric ratio should be used to find the hypotenuse?
$\tan$ only
$\sin$ or $\cos$
$\cot$ only
Any of $\sin$, $\cos$, $\tan$
practice An escalator in a shopping mall rises at an angle of $30^\circ$ to the horizontal floor. A shopper travels $8$ m along the escalator, from its bottom to its top. Find the vertical height gained.
practice A guy wire is stretched from the top of a mobile-network tower to a peg fixed in the ground, making an angle of $60^\circ$ with the ground. If the wire is $16$ m long, find the distance of the peg from the foot of the tower.
practice A ladder $8$ m long is placed so that it reaches a window $4$ m above the ground, at the foot of a vertical wall. What angle does the ladder make with the ground?
$30^\circ$
$45^\circ$
$60^\circ$
$90^\circ$
practice A person $1.6$ m tall stands $9$ m from the foot of a pole. The angle of elevation of the top of the pole, measured from the person’s eyes, is $60^\circ$. Find the straight-line distance from the person’s eyes to the top of the pole.
practice A person stands at some distance from a tower and works out, from the triangle formed by the angle of elevation, that the rise above eye level is $22$ m. If the person’s eyes are $1.5$ m above the ground, what is the height of the tower above the ground?
$20.5$ m
$22$ m
$23.5$ m
$33$ m
practice A flagstaff $7 \sqrt{3}$ m tall stands on level ground. From a point on the ground, the angle of elevation of its top is $60^\circ$. Find the distance of the point from the foot of the flagstaff.
practice From the top of a tower, the angle of depression of a car on the ground is $60^\circ$. What is the angle of elevation of the top of the tower, as seen from the car?
$30^\circ$
$45^\circ$
$60^\circ$
$90^\circ$
practice A fire engine’s ladder extends at an angle of $60^\circ$ to the ground to reach a window $15 \sqrt{3}$ m above the ground. Find the length the ladder must be extended to.
practice A helicopter is hovering at a height of $250$ m above a straight road. The angle of depression of a car on the road, measured from the helicopter, is $30^\circ$. Find the line-of-sight distance between the helicopter and the car.
practice A tree breaks in a storm partway up its trunk. The broken upper part bends over so that its tip touches the ground, making an angle of $30^\circ$ with the ground. The tip touches the ground $9$ m from the foot of the tree. Find the height of the tree before it broke.
practice A boy standing on level ground finds that the angle of elevation of the top of a building $45$ m tall increases from $30^\circ$ to $60^\circ$ as he walks in a straight line towards it. Find the distance he walked.
practice A tower and a building stand directly opposite each other on level ground. The angle of elevation of the top of the tower, measured from the foot of the building, is $60^\circ$. The angle of elevation of the top of the building, measured from the foot of the tower, is $30^\circ$. The tower is $30$ m tall. Find the height of the building.
practice From the top of a building $12$ m high, the angle of elevation of the top of a nearby cable tower is $60^\circ$, and the angle of depression of the tower’s foot is $45^\circ$. Find the height of the cable tower.
practice A girl $1.6$ m tall spots a balloon moving horizontally at a constant height of $61.6$ m above the ground. At one instant, the angle of elevation of the balloon from her eyes is $60^\circ$; a little later, it is $30^\circ$. Find the distance travelled by the balloon in that interval.
