Exam practice: Grade 10 Math Unit 15

a.

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

Answer

Angle of elevation: when an observer looks at an object higher than the eye, the angle between the line of sight and the horizontal line through the eye. Example: a student looking up at the top of a flagpole.

Angle of depression: when an observer looks at an object lower than the eye, the angle between the line of sight and the horizontal line through the eye. Example: a person on a roof looking down at a car on the road.

b.

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

Answer

tan: $\tan\theta = \dfrac{p}{b}$ .

c.

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

Answer

sin: $\sin\theta = \dfrac{p}{h}$ .

a.

Figure 15.15

Answer

$\tan 60° = \dfrac{AB}{BC}$ , so $\sqrt3 = \dfrac{5\sqrt3}{x}$ and x = 5 m.

b.

Figure 15.16

Answer

$\tan 30° = \dfrac{AB}{BC}$ , so $\dfrac{1}{\sqrt3} = \dfrac{x}{5\sqrt3}$ and x = 5.

c.

Figure 15.17

Answer

$\tan x° = \dfrac{AB}{BC} = \dfrac{10}{10} = 1 = \tan 45°$ , so x = 45.

a.

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

$\tan 30° = \dfrac{60}{x}$ , so $x = 60\sqrt3 \approx$ 103.92 m ( $60\sqrt3$ m).

b.

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

$\tan x° = \dfrac{12}{12} = 1$ , so x = 45.

c.

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

$\tan 45° = \dfrac{x}{12}$ , so x = 12 m.

a.

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

Figure 15.18

Let the broken part be x m; the standing part is (14 − x) m.

$\begin{aligned} \sin 60° &= \frac{14 - x}{x} \\ \frac{\sqrt3}{2}x &= 14 - x \\ x\left(1 + \frac{\sqrt3}{2}\right) &= 14 \\ x &= \frac{28}{2 + \sqrt3} \approx 7.50 \end{aligned}$

The broken part is about 7.5 m (6.5 m still stands).

b.

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

Standing part = $7.5 \sin 60° = 7.5 \times \dfrac{\sqrt3}{2} \approx 6.50$ m. Height = 7.5 + 6.50 ≈ 14 m (13.995 m).

c.

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

i.Standing part = $30 \sin 30° = 15$ m, so the tree was 30 + 15 = 45 m.
ii.Distance = $30 \cos 30° = 15\sqrt3 \approx$ 25.98 m ( $15\sqrt3$ m).

a.

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

Height above his eye = $30 \tan 60° = 30\sqrt3 \approx 51.96$ m. Tower = 51.96 + 1.7 ≈ 53.66 m.

b.

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

Height of the kite above his hand (eye) = $90\sqrt2 \sin 45° = 90\sqrt2 \times \dfrac{1}{\sqrt2} = 90$ m. Above the ground: 90 + 1.4 + 33.6 = 125 m. (Taking the string as straight, as the question means.)

c.

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

Height above his eye = 51.5 − 1.5 = 50 m. $\tan 45° = \dfrac{50}{d}$ , so d = 50 m.

a.

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

Height of the top above his eye = $20 \tan 60° = 20\sqrt3 \approx 34.64$ m. His eye = 36.5 − 34.64 ≈ 1.86 m.

b.

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

The roof is $10\sqrt3 \tan 30° = 10\sqrt3 \times \dfrac{1}{\sqrt3} = 10$ ft above the treetop. Tree = 30 − 10 = 20 ft.

c.

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

The tower’s top is $35 \tan 45° = 35$ m above the roof. House = 60 − 35 = 25 m.

a.

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

Radius = 45 m. $\tan\theta = \dfrac{45}{45} = 1$ , so $\theta = 45°$ .

b.

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

Radius = 65 m. Height = $65 \tan 45° =$ 65 m.

c.

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

Top above his eye = 11.62 − 1.62 = 10 m. Radius = $\dfrac{10}{\tan 60°} = \dfrac{10}{\sqrt3} \approx 5.77$ m. Diameter ≈ 11.55 m.

a.

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

$120 \sin 30° = 60$ m above his hand; kite = 60 + 1.5 = 61.5 m.

b.

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

Kite above his hand = 58 − 9 − 1.5 = 47.5 m. $\sin 30° = \dfrac{47.5}{L}$ , so L = 95: the string is 95 m long.

c.

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

$66\sqrt2 \sin 45° = 66$ m above his hand; kite = 66 + 2 + 32 = 100 m.

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

$\tan\theta = \dfrac{20}{20\sqrt3} = \dfrac{1}{\sqrt3}$ , so the sun is at 30°. The tower’s shadow = $\dfrac{25\sqrt3}{\tan 30°} = 25\sqrt3 \times \sqrt3 =$ 75 m.

a.

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

Top above his eye = 25 − 1.2 = 23.8 m. $\tan\theta = \dfrac{23.8}{23.8} = 1$ , so $\theta = 45°$ .

b.

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

Answer

Yes. With the distance fixed, a taller tree makes the perpendicular bigger, so $\tan\theta = \dfrac{p}{b}$ is bigger, and so is θ: he has to look up more steeply.

c.

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

Answer

It increases. The height stays the same but the base gets shorter, so $\tan\theta = \dfrac{p}{b}$ gets bigger, and so does θ.

a.

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

The angle of depression.

b.

Find that angle in degrees.

Answer

30°: it equals the cat’s angle of elevation (alternate angles between the two horizontals).

c.

How far is the cat from the pole?

Answer

$\tan 30° = \dfrac{2\sqrt3}{d}$ , so $d = 2\sqrt3 \times \sqrt3 =$ 6 m.

d.

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

Answer

No, it gets bigger. The height stays $2\sqrt3$ m while the distance shrinks, so tan θ grows and the parrot looks down more steeply.

a.

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

$2\pi r = 176$ , so $r = \dfrac{176 \times 7}{2 \times 22} = 28$ : 28 m.

b.

Find the height of the pole above the water.

Answer

Top above his eye = $28 \tan 45° = 28$ m. Pole = 28 + 1.6 = 29.6 m.

c.

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

Answer

For 30°: top above his eye = $\dfrac{28}{\sqrt3} \approx 16.17$ m, so the pole would be 16.17 + 1.6 = 17.77 m. Shorter by 29.6 − 17.77 ≈ 11.83 m.

a.

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

Figure 15.20 Drawn to scale: the man’s 1.2 m from the roof R to his eye M is too short to mark.

b.

How high is the kite above the ground?

Answer

$180 \sin 30° = 90$ m above his eye; kite = 90 + 1.2 + 8.8 = 100 m.

c.

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

Answer

$180 \cos 30° = 90\sqrt3 \approx$ 155.88 m ( $90\sqrt3$ m).

a.

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

An angle of depression: the shorter roof is below the eye.

b.

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

They are equal. It is the same line of sight PS, and the horizontals through P and S are parallel, so the two angles are alternate angles.

c.

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

Answer

ST = 32 − 20 = 12 m and PT = 12 m, so $\tan\theta = \dfrac{12}{12} = 1$ : 45°.

d.

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

Answer

$PS = \sqrt{12^2 + 12^2} = 12\sqrt2 \approx$ 16.97 m (or $\dfrac{12}{\sin 45°}$ ).

a.

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

$BD = 25 \tan 45° = 25$ m; tower = 25 + 15 = 40 m.

b.

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

Answer

It would increase: AD stays 25 m while BD grows, so tan θ grows.

a.

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

60°, equal to the angle of depression (alternate angles).

b.

Find the height of the part BC of the tower.

Answer

$BC = 20 \tan 60° = 20\sqrt3 \approx$ 34.64 m.

c.

Find the height of the house.

Answer

50 − 34.64 ≈ 15.36 m.

d.

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

At 45° the observer must be $20 \tan 45° = 20$ m above the roof, at 15.36 + 20 = 35.36 m. Come down 50 − 35.36 ≈ 14.64 m ( $20\sqrt3 - 20$ m).

Exam practice

Unit 15 · Trigonometry

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16