Trigonometry: Grade 10 Math – Science Playground

How tall is the Nyatapola? Nepali Times says 30 m; Wikipedia and the Bhaktapur.com guide say about 33 m. Three meters is a whole floor of a house, and nobody is going to climb the five roofs with a tape.

You don’t need to. Stand in Taumadhi Square, measure how far you are from the point under the top, and measure one angle: how far you tip your eyes up to see the finial. Trigonometry turns those two numbers into the height. This unit shows you how, with a protractor, a straw and a thread, and how to find any height or distance you can’t reach: a tree, a tower, a kite, the width of a pond.

Heights: Nepali Times, Wikipedia, bhaktapur.com (Sources).

The Nyatapola temple in Bhaktapur: five brick plinths with stone guardian figures on the stairs, and above them five dark tiered roofs, against a clear blue sky — **Figure 15.1** The Nyatapola in Taumadhi Square, Bhaktapur: five plinths, five roofs, built in 1702. Photo: Jean-Pierre Dalbéra, CC BY 2.0

Drag yourself (the yellow dot) along the ground. The temple is drawn 30 m tall. Watch the angle change, and the height the angle gives.

Distance d: 30.0 m
Angle θ: 43.5°
d × tan θ: 28.5 m
+ eye 1.5 m: 30.0 m

Wherever you stand, the angle changes but the height comes out the same: the temple as drawn is 30 m. Walk back and the angle falls; the distance grows to make up for it.

Rounded to a whole degree, the height is out by 0.0 m. Near 45° a small slip in the angle matters least.

Rounded to a whole degree, the height is out by 0.0 m. Close up the angle is steep, and a half degree’s slip moves the answer a lot.

For the SEE

The question: Trigonometry is 4 marks, one question in four parts of 1 mark each: knowledge (name an angle of elevation or depression), understanding (a ratio or a length), application (a height or distance) and higher ability (what changes when the observer moves, or a second step).
The rules it needs:

An angle of elevation is measured up from the horizontal through the eye; an angle of depression is measured down from it.
In a right-angled triangle: $\sin\theta = \dfrac{p}{h}$ , $\cos\theta = \dfrac{b}{h}$ , $\tan\theta = \dfrac{p}{b}$ . For heights and distances, tan is used most.
$\tan 30° = \dfrac{1}{\sqrt3}$ , $\tan 45° = 1$ , $\tan 60° = \sqrt3$ .
The angle of elevation from A to B equals the angle of depression from B to A (alternate angles).
Add the observer’s height (eye level) back on at the end.

How the SEE asks it has a full question marked part by part; Exam practice has every exercise question to mark yourself.

15.0 Review

Figure 15.2 Triangle ABC, right-angled at B, with the angle θ at C.

Grade 9 named the sides of a right-angled triangle from the angle you are looking at. For the angle θ at C in Figure 15.2:

the hypotenuse h is the longest side, opposite the right angle: AC;
the perpendicular p is the side opposite θ: AB;
the base b is the third side, next to θ: BC.

The three trigonometric ratios of θ are

$\sin\theta = \frac{p}{h} = \frac{AB}{AC}, \qquad \cos\theta = \frac{b}{h} = \frac{BC}{AC}, \qquad \tan\theta = \frac{p}{b} = \frac{AB}{BC}$

Why do these ratios belong to the angle? Every right-angled triangle with the angle θ at C is the same shape, only bigger or smaller (the triangles are similar), so its sides keep the same proportions. Double the triangle and AB and BC both double: tan θ doesn’t change. That is the whole trick of this unit: measure the angle, and you know the ratio of the sides, whatever their size.

θ	0°	30°	45°	60°	90°
sin θ	0	$\dfrac12$	$\dfrac{1}{\sqrt2}$	$\dfrac{\sqrt3}{2}$	1
cos θ	1	$\dfrac{\sqrt3}{2}$	$\dfrac{1}{\sqrt2}$	$\dfrac12$	0
tan θ	0	$\dfrac{1}{\sqrt3}$	1	$\sqrt3$	not defined

You don’t have to learn the table by heart: it comes from two set squares.

Figure 15.3 Half a square (left) and half an equilateral triangle (right). Their sides give every value in the table.

Cut a square with sides 1 along its diagonal: the diagonal is $\sqrt{1^2 + 1^2} = \sqrt2$ , and both acute angles are 45°. So $\tan 45° = \dfrac11 = 1$ and $\sin 45° = \cos 45° = \dfrac{1}{\sqrt2}$ . Cut an equilateral triangle with sides 2 down its middle: the half has angles 30° and 60°, a short side of 1 and an upright side of $\sqrt{2^2 - 1^2} = \sqrt3$ . So $\tan 60° = \sqrt3$ , $\tan 30° = \dfrac{1}{\sqrt3}$ , and the sines and cosines follow the same way.

In a right-angled triangle, which ratio uses only the perpendicular and the base?

Pick an answer to see why.

15.1 Angle of Elevation and Angle of Depression

Activity

1: how tall is that tree?

Ramnaresh is staring at a tree on the school ground when his mathematics teacher arrives.

Teacher: What is it, Ramnaresh?

Ramnaresh: Sir, how tall is that tree? How far am I from it? And how tall is the school’s flagpole? I want to find all of these without climbing anything.

Teacher: Then let’s go to the classroom and draw the question.

On the board the teacher draws Figure 15.4. Ramnaresh’s eye is at E, the tree is AB, and D is the point on the tree at the height of his eye, so BD = CE. If we can find AD, we add DB and have the tree.

Omkumari: What is the angle you make when you look up at the top called? How do we find it?

Teacher: First imagine a line from your eye, level with the ground: ED. When you look at the top, your line of sight EA makes an angle with that level line. That angle, ∠AED, is the angle of elevation.

Ramnaresh: And if someone at the top looks down at me?

Teacher: Imagine a level line from the top too, AF. The line of sight AE makes the angle ∠FAE with it, the angle of depression (Figure 15.5).

Figure 15.4 Looking up: the line of sight EA and the eye-level line ED make the angle of elevation θ.

Figure 15.5 Looking down from A: the level line AF and the line of sight AE make the angle of depression, the same θ.

When an observer looks at something higher than the eye, the angle between the line of sight and the horizontal line through the eye is the angle of elevation. When the observer looks at something lower than the eye, the angle between the line of sight and the horizontal is the angle of depression. Both are measured from the horizontal, never from the upright. An instrument that measures them is a clinometer.

In Figure 15.5, AF and DE are both horizontal, so they are parallel, and the line of sight AE cuts both. ∠FAE and ∠AED are alternate angles, so they are equal: the angle of depression from A to E equals the angle of elevation from E to A. One line of sight, one angle, seen from either end.

Top or middle? Ramnaresh asked one more thing: is the angle of elevation bigger when you look at the top of the tree or at its middle? Try it with your own eyes: you tip your head further back for the top. The higher the point you look at, the bigger the angle of elevation. And from the top of the tree looking down, the angle of depression is bigger for a point close to the tree than for one far away: the nearer the point, the more steeply you look down.

Sushma stands on the roof of her house and sees her friend on the road below. Which angle does she measure?

Pick an answer to see why.

From the top of a tower, a car on the road is seen at an angle of depression of 35°. From the car, what is the angle of elevation of the top of the tower?

Pick an answer to see why.

The clinometer

A Suunto clinometer: a flat metal case with a round window, through which a dial marked in degrees from 0 to 90 is seen — **Figure 15.6** A forester’s clinometer. You look through it at a treetop and read the angle of elevation on the dial. Photo: Btomli2, CC BY-SA 3.0

Foresters measure trees exactly the way this unit teaches: a clinometer for the angle to the top, a tape for the distance, and then “horizontal distance at eye level × tan θ = height above eye level”. That is called the tangent method. Its usual error is a top that isn’t straight above the base (a leaning tree, a branch reaching towards you): the method takes the top to be over the base, and the height almost always comes out too big.

You can make a clinometer that works on the same idea from a protractor.

15.2 Heights and distances

Every heights-and-distances problem goes the same way:

Draw the picture: the ground, the object, the observer, the dashed horizontal through the eye, the line of sight, the angle.
Find the right-angled triangle whose angle you know. Its right angle is where the horizontal through the eye meets the upright object.
Pick the ratio that joins the side you know and the side you want: opposite and adjacent, tan; opposite and hypotenuse, sin; adjacent and hypotenuse, cos.
Solve, then add or take away the eye height (or a building’s height) that isn’t inside the triangle.
Ask whether it makes sense: at 45° the height above the eye equals the distance; above 45° it is more, below 45° less.

Back to the Nyatapola

Yes, you can settle 30 m or 33 m from the square, with a homemade clinometer, if you are careful. Here is how big the difference is.

Say you stand 40 m from the point straight under the finial, with your eye 1.5 m above the paving. The top is 28.5 m above your eye if the temple is 30 m, or 31.5 m if it is 33 m:

$\begin{aligned} \tan\theta &= \frac{28.5}{40} = 0.7125, \text{ so } \theta \approx 35.5° \\ \tan\theta &= \frac{31.5}{40} = 0.7875, \text{ so } \theta \approx 38.2° \end{aligned}$

The two claims are about 2.7° apart. A clinometer read to the nearest degree tells them apart; one read carelessly, to within 2° or 3°, can’t. Three things matter:

The distance is to the point under the top, the middle of the temple, not to the bottom step. The five plinths spread far out under the roofs, so measure to the middle of the lowest plinth’s front and add half its width (pace it along the side).
Measure several times and take the mean: from two or three distances, each reading twice.
Stand back. Close up the angle is steep and a half-degree slip moves the answer a lot; near 45° it moves it least. Try “Clinometer reads whole degrees” in the widget at the top of this unit, near and far.

How the SEE asks it

CDC’s Specification Grid for Compulsory Mathematics gives Trigonometry 4 marks: one question with four parts, one each in knowledge, understanding, application and higher ability. The figure is usually given; the angles are 30°, 45° or 60°. Here is a question of our own in that shape.

Exercise 15

Question 1

Define the angle of elevation and the angle of depression, with an example of each.

Answer

In a right-angled triangle, which trigonometric ratio relates the perpendicular and the base?

Answer

In a right-angled triangle, which trigonometric ratio relates the perpendicular and the hypotenuse?

Answer

Question 2

Find the value of x in each right-angled triangle.

Figure 15.15

Answer

Figure 15.16

Answer

Figure 15.17

Answer

Question 3

A tower is 60 m high, and a man is x m from it. He sees its top at an angle of elevation of 30°. Find x. (Ignore his height.)

Answer

A tower is 12 m high and a man is 12 m from it. He sees its top at an angle of elevation of $x°$ . Find x.

Answer

A tower is x m high and a man is 12 m from it. He sees its top at an angle of elevation of 45°. Find x.

Answer

Question 4

A tree 14 m high breaks in the wind. Its top touches the ground, still joined to the trunk, and the broken part makes an angle of 60° with the ground. Find the length of the broken part.

Answer

A tree breaks partway up in the wind. Its top touches the ground, still joined to the trunk, making an angle of 60° with the ground. The broken part is 7.5 m long. How tall was the tree before it broke?

Answer

A tree breaks partway up in the wind. Its top touches the ground, still joined to the trunk, making an angle of 30° with the ground. The broken part is 30 m long.

i.How tall was the tree before it broke?
ii.How far from the foot of the tree does its top meet the ground?

Answer

Question 5

A man 1.7 m tall sees the top of a tower at an angle of elevation of 60°. He is 30 m from the tower. Find the height of the tower.

Answer

A man 1.4 m tall is flying a kite from the roof of a house 33.6 m high. The kite string is $90\sqrt2$ m long and makes an angle of 45° with the horizontal. Find the height of the kite above the ground.

Answer

A man 1.5 m tall sees the top of a tree 51.5 m high at an angle of elevation of 45°. Find the distance between the man and the tree.

Answer

Question 6

A man is 20 m from a tower 36.5 m high. The angle of elevation of its top from his eye is 60°. Find the height of his eye above the ground.

Answer

From the top of a house 30 ft high, the angle of depression of the top of a tree is 30°. The house and the tree are $10\sqrt3$ ft apart. Find the height of the tree.

Answer

A tower is 60 m high and a house is 35 m from it. From the top of the tower, the angle of depression of the top of the house is 45°. Find the height of the house.

Answer

Question 7

A circular pond is 90 m across. A pole at its center stands 45 m above the water. Find the angle of elevation of the top of the pole from the edge of the pond.

Answer

A circular pond is 130 m across. A pole stands at its center. From the edge, the angle of elevation of its top is 45°. Find the height of the pole above the water.

Answer

A pole stands at the center of a circular pond, 11.62 m above the water. From the edge of the pond, a man whose eye is 1.62 m above the water sees its top at an angle of elevation of 60°. Find the diameter of the pond.

Answer

Question 8

At a festival, Ramesh flies a kite. The string makes an angle of 30° with the horizontal and is 120 m long. Ramesh is 1.5 m tall. Find the height of the kite above the ground.

Answer

Ramsharan, 1.5 m tall, flies a kite from the roof of a house 9 m high. The string makes an angle of 30° with the horizontal, and the kite is 58 m above the ground. Find the length of the string.

Answer

A man 2 m tall flies a kite from the roof of a house 32 m high. The string is $66\sqrt2$ m long and makes an angle of 45° with the horizontal. Find the height of the kite above the ground.

Answer

Question 9

At 2 pm a pole 20 m high casts a shadow $20\sqrt3$ m long. Find the angle of elevation of the sun then, and the length of the shadow of a tower $25\sqrt3$ m high at the same time.

Answer

Question 10

A tree 25 m high stands at one corner of a school ground. A man whose eye is 1.2 m above the ground sits at another corner, 23.8 m from the tree.
What is the angle of elevation of the top of the tree from the man?

Answer

Would the angle of elevation increase if the tree were taller? Give a reason.

Answer

Does the angle of elevation increase or decrease if the man moves closer to the tree? Give a reason.

Answer

Question 11

A cat on the ground looks up at a parrot in a cage on a pole, $2\sqrt3$ m above the ground, at an angle of elevation of 30°.

Figure 15.19

What is the angle the parrot’s line of sight makes with the horizontal, when it looks down at the cat, called?

Answer

Find that angle in degrees.

Answer

How far is the cat from the pole?

Answer

As the cat moves towards the pole, does the angle the parrot makes looking at it get smaller? Give a reason.

Answer

Question 12

The circumference of a circular pond is 176 m. A pole stands at its center. From the edge of the pond, a man whose eye is 1.6 m above the water sees the top of the pole at an angle of elevation of 45°.
Find the distance between the man and the pole.

Answer

Find the height of the pole above the water.

Answer

How much shorter would the pole above the water have to be for the angle of elevation to be 30°?

Answer

Question 13

A man whose eye is 1.2 m above the roof flies a kite from the roof of a house 8.8 m high. The string is 180 m long and makes an angle of 30° with the horizontal.
Show the situation in a diagram.

Answer

How high is the kite above the ground?

Answer

What is the horizontal distance between the man and the kite?

Answer

Question 14

Two buildings, 20 m and 32 m high, stand 12 m apart.

Figure 15.21

Looking from the roof of the taller building to the roof of the shorter one, what kind of angle is formed?

Answer

How is the angle of elevation of the taller roof, seen from the shorter roof, related to the angle of depression of the shorter roof, seen from the taller one? Give a reason.

Answer

Find the angle of elevation of the taller roof from the shorter roof.

Answer

How long a ladder would reach from the top of the shorter building to the top of the taller one?

Answer

Question 15

A tower and a house stand 25 m apart on level ground. The house is 15 m high.

Figure 15.22

From a point A on the edge of the roof, the angle of elevation of the top B of the tower is 45°. Find the height of the tower.

Answer

How would the angle of elevation change if the tower were taller?

Answer

Question 16

From the top C of a view tower 50 m high, the angle of depression of a point A on the roof of a house 20 m to the east is 60°.

Figure 15.23

What is the angle of elevation of the top of the tower from A?

Answer

Find the height of the part BC of the tower.

Answer

Find the height of the house.

Answer

How far down from the top of the tower must an observer come to see the roof at A at an angle of depression of 45°?

Answer

Worksheet to print (PDF, 4 pages) Answer key for teachers

Harder worksheet (PDF, 4 pages) Answer key for teachers

Quiz yourself on this unit →

Practice every question in the book, and keep track of the ones you miss →

Where these facts come from

The Nyatapola’s height, 30 m: Nepali Times, “Nyatapola alight”, 12 May 2019, read October 2026. About 33 m (33.23 m), five plinths and five roofs, built 1701–1702: Wikipedia, “Nyatapola”, read October 2026; “the ultimate height of the temple is 33 meters”: bhaktapur.com, “Nyatapola Temple”, read October 2026.
The tangent method (clinometer and tape; “horizontal distance at eye level × tangent Θ = height above eye level”; a top not over the base adds to the height): Wikipedia, “Tree height measurement”, read October 2026.
Thales measuring the pyramids by their shadows when his shadow equaled his height, as told by Diogenes Laërtius, Pliny and Plutarch from Hieronymus of Rhodes (the story): Wikipedia, “Thales of Miletus”, read October 2026.
Peak XV observed from stations up to 240 km away, given 8,840 m (29,002 ft) in 1856, refraction, the 2020 height 8,848.86 m (fact box): Wikipedia, “Mount Everest”, read October 2026.
The glide slope about 3° above the horizontal (fact box): Wikipedia, “Instrument landing system”, read October 2026.
The SEE’s Trigonometry marks, one 4-mark question with parts K1 U1 A1 HA1 (For the SEE, How the SEE asks it): CDC, Specification Grid, Compulsory Mathematics, Grades 9–10 (2078), the grid table.
Photos: Wikimedia Commons, credited under each.

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