Class 10 Mathematics Chapter 9 Question Answer | Some Applications of Trigonometry | English Medium

Class 10 Mathematics Chapter 9 Question Answer | Some Applications of Trigonometry | English Medium | ASSEB

Welcome to HSLC Guru, dear students. This page brings you a complete, exam-ready guide to Class 10 Mathematics Chapter 9 — Some Applications of Trigonometry, prepared in English medium and aligned with the ASSEB (Assam State School Education Board) syllabus. In this chapter we leave behind abstract triangles and step into the real world: the heights of towers, the depths of rivers, the lengths of ladders, the strings of kites, the depression of ships at sea and the elevation of balloons in the sky. Every solution below is presented with a clear right-triangle figure, the correct trigonometric ratio, full working in KaTeX, and the final numerical answer.

Trigonometry was developed centuries ago precisely to answer questions that could not be measured directly — How tall is that mountain? How far is that ship? How wide is that canal? In Chapter 9 you will learn that with just one angle and one length, you can compute almost any unknown distance using the simple relations $\tan\theta$, $\sin\theta$ and $\cos\theta$. Let us begin.

Chapter Summary

This chapter teaches us how to use the trigonometric ratios studied in Chapter 8 to solve practical problems involving heights and distances. We model a real situation as a right-angled triangle, identify the angle of elevation or depression, and apply $\sin$, $\cos$ or $\tan$ to find the required side. Exercise 9.1 contains 16 questions covering ladders, broken trees, slides, towers, kites, buildings, statues on pedestals, two-pole problems, lighthouses, balloons and moving cars.

Topic	Key Idea
Line of Sight	Straight line from observer’s eye to the object viewed
Horizontal Line	Reference line parallel to the ground passing through observer’s eye
Angle of Elevation	Angle between line of sight and horizontal when object is above
Angle of Depression	Angle between line of sight and horizontal when object is below
Main Tool	Right triangle with $\tan\theta = \dfrac{\text{opposite}}{\text{adjacent}}$
Useful Values	$\tan 30^\circ = \dfrac{1}{\sqrt{3}},\ \tan 45^\circ = 1,\ \tan 60^\circ = \sqrt{3}$

Key Concepts and Definitions

1. Line of Sight

The line of sight is the straight line drawn from the eye of an observer to the point in the object being viewed. When you look at the top of a tower, the line joining your eye and the tower-top is the line of sight.

2. Horizontal Level

The horizontal line is the line parallel to the ground passing through the observer’s eye. It is the natural reference for measuring how steeply the line of sight tilts upward or downward.

3. Angle of Elevation

When the object viewed is above the horizontal level, the angle formed between the line of sight and the horizontal is called the angle of elevation. The eye must be raised to view the object, so the angle opens upward.

4. Angle of Depression

When the object viewed is below the horizontal level, the angle formed between the horizontal and the line of sight is the angle of depression. Here the eye must look downward, so the angle opens below the horizontal.

5. Working Method

Read the question and draw a labelled right-triangle figure.
Mark the angle of elevation or depression at the observer.
Identify which side is opposite, adjacent and hypotenuse to the angle.
Choose the trigonometric ratio that involves the known side and the unknown side.
Substitute standard values of $\sin$, $\cos$, $\tan$ at $30^\circ, 45^\circ, 60^\circ$ and simplify.

Standard values to memorise:

$$\tan 30^\circ = \tfrac{1}{\sqrt 3}, \quad \tan 45^\circ = 1, \quad \tan 60^\circ = \sqrt 3$$

$$\sin 30^\circ = \tfrac{1}{2}, \quad \sin 45^\circ = \tfrac{1}{\sqrt 2}, \quad \sin 60^\circ = \tfrac{\sqrt 3}{2}$$

$$\cos 30^\circ = \tfrac{\sqrt 3}{2}, \quad \cos 45^\circ = \tfrac{1}{\sqrt 2}, \quad \cos 60^\circ = \tfrac{1}{2}$$

Exercise 9.1 — Complete Solutions

Question 1

A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30°.

Answer: Let AB be the pole of height $h$ and AC be the rope. Then $\angle ACB = 30^\circ$ and $AC = 20$ m.

In the right triangle ABC,

$$\sin 30^\circ = \frac{AB}{AC} = \frac{h}{20}$$

$$\frac{1}{2} = \frac{h}{20} \implies h = 10$$

Therefore, the height of the pole is 10 m.

Question 2

A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.

Answer: Let AB be the original tree. After the storm, it breaks at C; the broken part CB’ touches the ground at B’ with $\angle CB’A = 30^\circ$ and $AB’ = 8$ m.

From right triangle CAB’,

$$\tan 30^\circ = \frac{AC}{AB’} \implies \frac{1}{\sqrt 3} = \frac{AC}{8} \implies AC = \frac{8}{\sqrt 3}\text{ m}$$

$$\cos 30^\circ = \frac{AB’}{CB’} \implies \frac{\sqrt 3}{2} = \frac{8}{CB’} \implies CB’ = \frac{16}{\sqrt 3}\text{ m}$$

Total height $= AC + CB’ = \dfrac{8}{\sqrt 3} + \dfrac{16}{\sqrt 3} = \dfrac{24}{\sqrt 3} = 8\sqrt 3$ m.

Therefore, the height of the tree is $8\sqrt 3$ m.

Question 3

A contractor plans to install two slides for the children to play in a park. For children below 5 years she prefers a slide whose top is at a height of 1.5 m and inclined at 30° to the ground, while for elder children she wants a steeper slide at a height of 3 m and inclined at 60° to the ground. What should be the length of the slide in each case?

Answer: Let $L_1$ and $L_2$ be the lengths of the two slides.

For the younger children’s slide:

$$\sin 30^\circ = \frac{1.5}{L_1} \implies \frac{1}{2} = \frac{1.5}{L_1} \implies L_1 = 3\text{ m}$$

For the elder children’s slide:

$$\sin 60^\circ = \frac{3}{L_2} \implies \frac{\sqrt 3}{2} = \frac{3}{L_2} \implies L_2 = \frac{6}{\sqrt 3} = 2\sqrt 3\text{ m}$$

Hence the slides are 3 m and $2\sqrt 3$ m long respectively.

Question 4

The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.

Answer: Let $h$ be the height of the tower. From the right triangle formed,

$$\tan 30^\circ = \frac{h}{30} \implies \frac{1}{\sqrt 3} = \frac{h}{30} \implies h = \frac{30}{\sqrt 3} = 10\sqrt 3$$

Therefore, the height of the tower is $10\sqrt 3$ m $\approx 17.32$ m.

Question 5

A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.

Answer: Let $L$ be the length of the string. The vertical height is 60 m.

$$\sin 60^\circ = \frac{60}{L} \implies \frac{\sqrt 3}{2} = \frac{60}{L} \implies L = \frac{120}{\sqrt 3} = 40\sqrt 3$$

Therefore, the length of the string is $40\sqrt 3$ m.

Question 6

A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

Answer: Let the building be AB = 30 m. The boy’s eye level is 1.5 m, so the effective vertical for the right triangles is $30 – 1.5 = 28.5$ m.

Let the boy initially be at C and finally at D, with foot of building at E so that the horizontal eye-level meets the building at A’. Let $A’D = x$ and $CD = y$ (the distance walked).

From point D ($60^\circ$):

$$\tan 60^\circ = \frac{28.5}{x} \implies \sqrt 3 = \frac{28.5}{x} \implies x = \frac{28.5}{\sqrt 3} = \frac{28.5\sqrt 3}{3}$$

From point C ($30^\circ$):

$$\tan 30^\circ = \frac{28.5}{x + y} \implies \frac{1}{\sqrt 3} = \frac{28.5}{x + y} \implies x + y = 28.5\sqrt 3$$

Subtracting,

$$y = 28.5\sqrt 3 – \frac{28.5\sqrt 3}{3} = \frac{2 \times 28.5\sqrt 3}{3} = \frac{57\sqrt 3}{3} = 19\sqrt 3$$

Therefore, the distance walked towards the building is $19\sqrt 3$ m $\approx 32.9$ m.

Question 7

From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.

Answer: Let AB = 20 m be the building and BC = $h$ be the tower fixed on top. Let D be the observation point at distance $d$ from A.

From the lower triangle ($45^\circ$ to top of building):

$$\tan 45^\circ = \frac{20}{d} \implies 1 = \frac{20}{d} \implies d = 20\text{ m}$$

From the larger triangle ($60^\circ$ to top of tower):

$$\tan 60^\circ = \frac{20 + h}{d} = \frac{20 + h}{20}$$

$$\sqrt 3 = \frac{20 + h}{20} \implies 20 + h = 20\sqrt 3 \implies h = 20(\sqrt 3 – 1)$$

Therefore, the height of the transmission tower is $20(\sqrt 3 – 1)$ m $\approx 14.64$ m.

Question 8

A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.

Answer: Let the pedestal height be $h$ and distance from observer = $d$.

From $45^\circ$ to top of pedestal:

$$\tan 45^\circ = \frac{h}{d} \implies d = h$$

From $60^\circ$ to top of statue (height $h + 1.6$):

$$\tan 60^\circ = \frac{h + 1.6}{d} = \frac{h + 1.6}{h}$$

$$\sqrt 3 \cdot h = h + 1.6 \implies h(\sqrt 3 – 1) = 1.6$$

$$h = \frac{1.6}{\sqrt 3 – 1} = \frac{1.6(\sqrt 3 + 1)}{(\sqrt 3 – 1)(\sqrt 3 + 1)} = \frac{1.6(\sqrt 3 + 1)}{2} = 0.8(\sqrt 3 + 1)$$

Therefore, the height of the pedestal is $0.8(\sqrt 3 + 1)$ m $\approx 2.19$ m.

Question 9

The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

Answer: Let height of building = $h$, distance between foot of building and foot of tower = $d$.

From foot of building, looking at tower top ($60^\circ$):

$$\tan 60^\circ = \frac{50}{d} \implies \sqrt 3 = \frac{50}{d} \implies d = \frac{50}{\sqrt 3}$$

From foot of tower, looking at building top ($30^\circ$):

$$\tan 30^\circ = \frac{h}{d} \implies \frac{1}{\sqrt 3} = \frac{h}{50/\sqrt 3} \implies h = \frac{50}{\sqrt 3 \cdot \sqrt 3} = \frac{50}{3}$$

Therefore, the height of the building is $\dfrac{50}{3}$ m $\approx 16.67$ m.

Question 10

Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30° respectively. Find the height of the poles and the distances of the point from the poles.

Answer: Let the height of each pole = $h$. Let P be the observation point, with distance $x$ from the first pole and $(80 – x)$ from the second pole.

From the first pole ($60^\circ$):

$$\tan 60^\circ = \frac{h}{x} \implies h = x\sqrt 3$$

From the second pole ($30^\circ$):

$$\tan 30^\circ = \frac{h}{80 – x} \implies h = \frac{80 – x}{\sqrt 3}$$

Equating,

$$x\sqrt 3 = \frac{80 – x}{\sqrt 3} \implies 3x = 80 – x \implies 4x = 80 \implies x = 20$$

So $h = 20\sqrt 3$ m and the other distance $= 80 – 20 = 60$ m.

Therefore, the height of each pole is $20\sqrt 3$ m and the point is at distances of 20 m and 60 m from the two poles.

Question 11

A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the canal.

Answer: Let the canal width = $x$ m and tower height = $h$ m. Point B is on the opposite bank, point A is $20$ m further away.

From B ($60^\circ$):

$$\tan 60^\circ = \frac{h}{x} \implies h = x\sqrt 3 \quad\ldots(i)$$

From A ($30^\circ$):

$$\tan 30^\circ = \frac{h}{x + 20} \implies \frac{1}{\sqrt 3} = \frac{h}{x + 20}$$

$$h\sqrt 3 = x + 20 \quad\ldots(ii)$$

From (i), substitute $h = x\sqrt 3$ in (ii):

$$(x\sqrt 3)\sqrt 3 = x + 20 \implies 3x = x + 20 \implies x = 10$$

So $h = 10\sqrt 3$ m.

Therefore, the height of the tower is $10\sqrt 3$ m and the width of the canal is 10 m.

Question 12

From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

Answer: Let the cable tower be CD with foot D and top C. Let the building be AB with top A. Draw AE perpendicular to CD where $E$ is on CD with $AE \parallel BD$. Then $BD = AE = x$ and $ED = AB = 7$ m.

From angle of depression $45^\circ$ to D:

$$\tan 45^\circ = \frac{7}{x} \implies x = 7\text{ m}$$

From angle of elevation $60^\circ$ to C:

$$\tan 60^\circ = \frac{CE}{x} \implies CE = 7\sqrt 3\text{ m}$$

Total tower height $= CE + ED = 7\sqrt 3 + 7 = 7(\sqrt 3 + 1)$ m.

Therefore, the height of the cable tower is $7(\sqrt 3 + 1)$ m $\approx 19.12$ m.

Question 13

As observed from the top of a 75 m high lighthouse from the sea level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Answer: Let the lighthouse be AB = 75 m, with foot at B. Let the nearer ship be at C ($45^\circ$) and the farther ship at D ($30^\circ$).

From angle of depression $45^\circ$:

$$\tan 45^\circ = \frac{75}{BC} \implies BC = 75\text{ m}$$

From angle of depression $30^\circ$:

$$\tan 30^\circ = \frac{75}{BD} \implies \frac{1}{\sqrt 3} = \frac{75}{BD} \implies BD = 75\sqrt 3\text{ m}$$

Distance between ships $= BD – BC = 75\sqrt 3 – 75 = 75(\sqrt 3 – 1)$ m.

Therefore, the distance between the two ships is $75(\sqrt 3 – 1)$ m $\approx 54.9$ m.

Question 14

A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30°. Find the distance travelled by the balloon during the interval.

Answer: The girl’s eye level is 1.2 m above the ground, so the vertical height of the balloon above her eye is $88.2 – 1.2 = 87$ m.

Let the horizontal distances at the two instants be $x_1$ (at $60^\circ$) and $x_2$ (at $30^\circ$).

$$\tan 60^\circ = \frac{87}{x_1} \implies x_1 = \frac{87}{\sqrt 3} = 29\sqrt 3$$

$$\tan 30^\circ = \frac{87}{x_2} \implies x_2 = 87\sqrt 3$$

Distance travelled by the balloon $= x_2 – x_1 = 87\sqrt 3 – 29\sqrt 3 = 58\sqrt 3$ m.

Therefore, the balloon travelled $58\sqrt 3$ m.

Question 15

A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

Answer: Let height of tower = $h$, distance of car at $30^\circ$ position = $x$ (initial), distance at $60^\circ$ position = $y$ (after 6 s).

$$\tan 30^\circ = \frac{h}{x} \implies x = h\sqrt 3$$

$$\tan 60^\circ = \frac{h}{y} \implies y = \frac{h}{\sqrt 3}$$

Distance travelled in 6 s $= x – y = h\sqrt 3 – \dfrac{h}{\sqrt 3} = \dfrac{3h – h}{\sqrt 3} = \dfrac{2h}{\sqrt 3}$.

Speed of car $= \dfrac{2h}{6\sqrt 3} = \dfrac{h}{3\sqrt 3}$ m/s.

Time to cover remaining distance $y = \dfrac{h}{\sqrt 3}$:

$$\text{time} = \frac{h/\sqrt 3}{h/(3\sqrt 3)} = \frac{h}{\sqrt 3} \times \frac{3\sqrt 3}{h} = 3\text{ seconds}$$

Therefore, the car takes 3 seconds more to reach the foot of the tower.

Question 16

The angles of elevation of the top of a tower from two points at distances of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.

Answer: Let the height of the tower be $h$. Let the angle of elevation from the point at 9 m be $\theta$; since the angles are complementary, the angle from the point at 4 m is $(90^\circ – \theta)$.

From the point at 9 m:

$$\tan\theta = \frac{h}{9}$$

From the point at 4 m:

$$\tan(90^\circ – \theta) = \cot\theta = \frac{h}{4}$$

Multiplying the two relations:

$$\tan\theta \cdot \cot\theta = \frac{h}{9} \cdot \frac{h}{4}$$

$$1 = \frac{h^2}{36} \implies h^2 = 36 \implies h = 6$$

Hence the height of the tower is 6 m. Proved.

Additional Practice Questions

AP-1

The angle of elevation of the top of a tower from a point on the ground is 30°. On walking 24 m towards the tower, the angle of elevation becomes 60°. Find the height of the tower.

Answer: Let height = $h$, distance of nearer point from foot = $x$.

$$\tan 60^\circ = \frac{h}{x} \implies x = \frac{h}{\sqrt 3}$$

$$\tan 30^\circ = \frac{h}{x + 24} \implies x + 24 = h\sqrt 3$$

Substituting,

$$\frac{h}{\sqrt 3} + 24 = h\sqrt 3 \implies h\sqrt 3 – \frac{h}{\sqrt 3} = 24 \implies \frac{2h}{\sqrt 3} = 24$$

$$h = 12\sqrt 3$$

The height of the tower is $12\sqrt 3$ m $\approx 20.78$ m.

AP-2

A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from the base of the wall and the angle the ladder makes with the ground.

Answer: Let the distance from the wall be $x$ and angle = $\theta$. Hypotenuse = 10 m, opposite = 8 m.

$$x = \sqrt{10^2 – 8^2} = \sqrt{36} = 6\text{ m}$$

$$\sin\theta = \frac{8}{10} = 0.8 \implies \theta \approx 53.13^\circ$$

Foot of ladder is 6 m from the wall and the angle is approximately $53.13^\circ$.

AP-3

The shadow of a tower standing on level ground is 30 m longer when the sun’s altitude is 30° than when it is 60°. Find the height of the tower.

Answer: Let height = $h$, shadow at $60^\circ$ = $x$, then shadow at $30^\circ$ = $x + 30$.

$$\tan 60^\circ = \frac{h}{x} \implies x = \frac{h}{\sqrt 3}$$

$$\tan 30^\circ = \frac{h}{x + 30} \implies x + 30 = h\sqrt 3$$

$$h\sqrt 3 – \frac{h}{\sqrt 3} = 30 \implies \frac{2h}{\sqrt 3} = 30 \implies h = 15\sqrt 3$$

Height of the tower is $15\sqrt 3$ m $\approx 25.98$ m.

AP-4

From a window 10 m above the ground in a street, the angles of elevation and depression of the top and the foot of another house on the opposite side of the street are 60° and 45° respectively. Find the height of the opposite house.

Answer: Let width of street = $x$, height of opposite house above window = $y$.

From depression $45^\circ$ to foot:

$$\tan 45^\circ = \frac{10}{x} \implies x = 10\text{ m}$$

From elevation $60^\circ$ to top:

$$\tan 60^\circ = \frac{y}{10} \implies y = 10\sqrt 3$$

Total height of opposite house $= 10 + 10\sqrt 3 = 10(1 + \sqrt 3)$ m $\approx 27.32$ m.

AP-5

An aeroplane flying horizontally at a height of 1500 m above the ground is observed at an elevation of 60°. After 15 seconds the elevation is observed to be 30°. Find the speed of the plane in km/h.

Answer: Let horizontal distances be $x_1$ (at $60^\circ$) and $x_2$ (at $30^\circ$).

$$x_1 = \frac{1500}{\sqrt 3} = 500\sqrt 3, \quad x_2 = 1500\sqrt 3$$

Distance travelled $= 1500\sqrt 3 – 500\sqrt 3 = 1000\sqrt 3$ m in 15 s.

$$\text{Speed} = \frac{1000\sqrt 3}{15} = \frac{200\sqrt 3}{3}\text{ m/s}$$

$$= \frac{200\sqrt 3}{3} \times \frac{3600}{1000} = 240\sqrt 3 \text{ km/h} \approx 415.69\text{ km/h}$$

The aeroplane’s speed is $240\sqrt 3$ km/h.

AP-6

The angle of elevation of the top of a chimney from the foot of a tower is 60° and the angle of depression from the top of the tower of the foot of the chimney is 30°. If the height of the tower is 40 m, find the height of the chimney. According to pollution control norms, the minimum height of the chimney must be 100 m. State if the height of the chimney meets the pollution norms. (Take $\sqrt 3 = 1.732$.)

Answer: Let the distance between the tower and chimney be $d$. The tower is 40 m high.

From angle of depression $30^\circ$ from top of tower:

$$\tan 30^\circ = \frac{40}{d} \implies d = 40\sqrt 3$$

From angle of elevation $60^\circ$ from foot of tower to top of chimney of height $h$:

$$\tan 60^\circ = \frac{h}{d} \implies h = d\sqrt 3 = 40\sqrt 3 \cdot \sqrt 3 = 120$$

The height of the chimney is 120 m. Since $120 > 100$, the chimney meets the pollution norms.

AP-7

From a point 40 m away from the foot of a tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower.

Answer:

$$\tan 30^\circ = \frac{h}{40} \implies h = \frac{40}{\sqrt 3} = \frac{40\sqrt 3}{3}\text{ m} \approx 23.09\text{ m}$$

Height of the tower $\approx$ $\dfrac{40\sqrt 3}{3}$ m.

AP-8

From a point P on the ground, the angle of elevation of the top of a 10 m tall building is 30°. A flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff from P is 45°. Find the length of the flagstaff and the distance of the building from the point P. (Take $\sqrt 3 = 1.732$.)

Answer: Let the distance from P to the foot of the building be $d$ and the length of the flagstaff be $\ell$.

From elevation $30^\circ$ to the top of building (10 m):

$$\tan 30^\circ = \frac{10}{d} \implies d = 10\sqrt 3 \approx 17.32\text{ m}$$

From elevation $45^\circ$ to the top of flagstaff (height $10 + \ell$):

$$\tan 45^\circ = \frac{10 + \ell}{10\sqrt 3} \implies 10 + \ell = 10\sqrt 3 \implies \ell = 10(\sqrt 3 – 1) \approx 7.32$$

Length of flagstaff $\approx$ 7.32 m; distance of building from P $\approx$ 17.32 m.

AP-9

The shadow of a vertical pole becomes 30 m longer when the sun’s altitude changes from 60° to 45°. Find the height of the pole.

Answer: Let the height be $h$. The shorter shadow corresponds to higher altitude.

$$\tan 60^\circ = \frac{h}{x_1} \implies x_1 = \frac{h}{\sqrt 3}$$

$$\tan 45^\circ = \frac{h}{x_1 + 30} \implies x_1 + 30 = h$$

$$\frac{h}{\sqrt 3} + 30 = h \implies h\left(1 – \frac{1}{\sqrt 3}\right) = 30$$

$$h \cdot \frac{\sqrt 3 – 1}{\sqrt 3} = 30 \implies h = \frac{30\sqrt 3}{\sqrt 3 – 1} = \frac{30\sqrt 3(\sqrt 3 + 1)}{2} = 15\sqrt 3(\sqrt 3 + 1) = 45 + 15\sqrt 3$$

Height of pole $= 15(3 + \sqrt 3)$ m $\approx 70.98$ m.

AP-10

A man on the deck of a ship 16 m above water level observes that the angle of elevation of the top of a cliff is 45° and the angle of depression of the base of the cliff is 30°. Find the height of the cliff and the distance of the cliff from the ship.

Answer: Let horizontal distance from ship to cliff = $d$, height of cliff above water = $H$.

Depression $30^\circ$ to base:

$$\tan 30^\circ = \frac{16}{d} \implies d = 16\sqrt 3$$

Elevation $45^\circ$ to top — height above deck = $H – 16$:

$$\tan 45^\circ = \frac{H – 16}{d} \implies H – 16 = 16\sqrt 3 \implies H = 16(1 + \sqrt 3)$$

Cliff height $= 16(1 + \sqrt 3) \approx 43.71$ m; distance from ship $= 16\sqrt 3 \approx 27.71$ m.

AP-11

A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height 5 m. From a point on the plane, the angles of elevation of the bottom and the top of the flagstaff are respectively 30° and 60°. Find the height of the tower.

Answer: Let height of tower = $h$, distance of point from foot = $d$.

$$\tan 30^\circ = \frac{h}{d} \implies d = h\sqrt 3$$

$$\tan 60^\circ = \frac{h + 5}{d} = \frac{h + 5}{h\sqrt 3}$$

$$\sqrt 3 \cdot h\sqrt 3 = h + 5 \implies 3h = h + 5 \implies h = \frac{5}{2} = 2.5\text{ m}$$

Height of the tower is 2.5 m.

AP-12

The angles of depression of the top and bottom of an 8 m tall building from the top of a multi-storeyed building are 30° and 45° respectively. Find the height of the multi-storeyed building and the distance between the two buildings.

Answer: Let multi-storeyed building height = $H$, distance between buildings = $d$. Smaller building = 8 m.

Depression $45^\circ$ to bottom of small building:

$$\tan 45^\circ = \frac{H}{d} \implies d = H$$

Depression $30^\circ$ to top of small building (vertical drop = $H – 8$):

$$\tan 30^\circ = \frac{H – 8}{d} = \frac{H – 8}{H} \implies \frac{1}{\sqrt 3} = \frac{H – 8}{H}$$

$$H = (H – 8)\sqrt 3 \implies H(\sqrt 3 – 1) = 8\sqrt 3$$

$$H = \frac{8\sqrt 3}{\sqrt 3 – 1} = \frac{8\sqrt 3(\sqrt 3 + 1)}{2} = 4\sqrt 3(\sqrt 3 + 1) = 12 + 4\sqrt 3$$

Height $\approx$ $4(3 + \sqrt 3)$ m $\approx 18.93$ m; distance same value $\approx 18.93$ m.

Important Tips for Solving Heights and Distances

Always draw a clear figure. Mark the right angle, the angle of elevation/depression, and label all known and unknown sides.
Identify the right triangle correctly. The vertical is usually the object’s height (tower, pole, building); the horizontal is the ground distance.
Choose the correct ratio. If you know hypotenuse and want vertical, use $\sin$. If you know horizontal and want vertical, use $\tan$. If you know hypotenuse and want horizontal, use $\cos$.
Mind the observer’s height. When an observer is taller than zero (boy 1.5 m, girl 1.2 m), subtract this from the object’s height to get the vertical of the right triangle.
Angle of depression at observer = Angle of elevation at object. Alternate angles are equal because the horizontal at the top is parallel to the ground.
Rationalise denominators. Convert $\dfrac{30}{\sqrt 3}$ to $10\sqrt 3$ for cleaner final answers.
Use complementary-angle property ($\tan\theta \cdot \cot\theta = 1$) when angles in a problem add to $90^\circ$.
Check units. If the answer asks for km/h, convert m/s by multiplying with $\dfrac{18}{5}$.

Quick Revision Sheet

Situation	Triangle Set-up	Useful Ratio
Tower of height $h$, point at distance $d$, elevation $\theta$	Vertical $= h$, horizontal $= d$	$\tan\theta = h/d$
Ladder of length $L$ leaning against wall, foot at distance $b$, top at height $h$	Hypotenuse $= L$	$\sin\theta = h/L,\ \cos\theta = b/L$
Kite at height $h$, string $L$ at inclination $\theta$	Hypotenuse $= L$	$\sin\theta = h/L$
Lighthouse of height $H$, ship at depression $\theta$, distance $d$	Vertical $= H$, horizontal $= d$	$\tan\theta = H/d$
Two angles of elevation $\alpha, \beta$ from points $a, b$ apart	Two right triangles, common height $h$	Set up two $\tan$ equations and subtract
Complementary angles from distances $a, b$	$\tan\theta = h/b,\ \cot\theta = h/a$	$h^2 = ab \implies h = \sqrt{ab}$

The complementary-angle shortcut is especially useful: if angles of elevation from points $a$ and $b$ from the foot of a vertical pole are complementary, then the height of the pole is $\sqrt{ab}$. For Question 16 of the exercise, $\sqrt{4 \times 9} = 6$ m, exactly as proved.

Glossary

Term	Meaning
Line of Sight	The straight line from observer’s eye to the point on the object viewed
Horizontal Line	Line through observer’s eye parallel to the ground
Angle of Elevation	Acute angle measured upward from horizontal to line of sight (object above)
Angle of Depression	Acute angle measured downward from horizontal to line of sight (object below)
Theodolite	Surveyor’s instrument used to measure angles of elevation and depression
Hypotenuse	Side opposite to the right angle, longest side of a right triangle
Opposite Side	Side opposite to the angle being considered
Adjacent Side	Side next to the angle being considered (not the hypotenuse)
Complementary Angles	Two angles whose sum is $90^\circ$; their tan and cot are reciprocals
Sine ($\sin\theta$)	Ratio $\dfrac{\text{opposite}}{\text{hypotenuse}}$
Cosine ($\cos\theta$)	Ratio $\dfrac{\text{adjacent}}{\text{hypotenuse}}$
Tangent ($\tan\theta$)	Ratio $\dfrac{\text{opposite}}{\text{adjacent}}$
Standard Angles	$0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ$ — the angles for which exact ratios are memorised
Rationalisation	Removing surds from a denominator, e.g. $\dfrac{1}{\sqrt 3} = \dfrac{\sqrt 3}{3}$
Uniform Speed	Constant speed; distance $=$ speed $\times$ time

Extended Worked Examples

Worked Example 1 — A Surveyor on a Hill

A surveyor stands on the top of a 60 m hill. From this point, he observes the angle of depression of two milestones on the road on the same side as 30° and 45°. Find the distance between the two milestones.

Solution. Let the foot of the hill be B, the top be A. Let nearer milestone $M_1$ ($45^\circ$) be at horizontal distance $x$ and farther milestone $M_2$ ($30^\circ$) at distance $y$.

$$\tan 45^\circ = \frac{60}{x} \implies x = 60\text{ m}$$

$$\tan 30^\circ = \frac{60}{y} \implies y = 60\sqrt 3\text{ m}$$

Distance between milestones $= y – x = 60\sqrt 3 – 60 = 60(\sqrt 3 – 1) \approx 43.92$ m.

Worked Example 2 — Angles in Both Directions

From the top of a tower 30 m high, the angle of depression of a man standing at the foot of an electric pole on the same side is 60°. From the foot of the tower, the angle of elevation of the top of the pole is 30°. Find the height of the pole.

Solution. Let pole height = $h$, distance between tower foot and pole foot = $d$.

From depression $60^\circ$ to man:

$$\tan 60^\circ = \frac{30}{d} \implies d = \frac{30}{\sqrt 3} = 10\sqrt 3\text{ m}$$

From foot of tower, elevation $30^\circ$:

$$\tan 30^\circ = \frac{h}{d} = \frac{h}{10\sqrt 3} \implies h = 10\sqrt 3 \cdot \frac{1}{\sqrt 3} = 10\text{ m}$$

The pole is 10 m high.

Worked Example 3 — Distance Between Two Vehicles

A jet plane flying at a height of 3000 m horizontally observes the angles of depression of two tanks on the ground as 30° and 60°. If the tanks are on the opposite sides of the plane, find the distance between them.

Solution. Let the foot of the perpendicular from the plane to the ground be P. Let $T_1, T_2$ be the tanks at horizontal distances $x_1$ and $x_2$ from P.

$$\tan 60^\circ = \frac{3000}{x_1} \implies x_1 = \frac{3000}{\sqrt 3} = 1000\sqrt 3$$

$$\tan 30^\circ = \frac{3000}{x_2} \implies x_2 = 3000\sqrt 3$$

Distance between tanks $= x_1 + x_2 = 1000\sqrt 3 + 3000\sqrt 3 = 4000\sqrt 3$ m $\approx 6928$ m.

Worked Example 4 — Two Towers

Two pillars of equal height stand on either side of a roadway 150 m wide. From a point on the roadway between the pillars, the angles of elevation of the tops of the pillars are 60° and 30°. Find the height of each pillar and the position of the point.

Solution. Let height = $h$, distance from $60^\circ$ pillar = $x$, distance from $30^\circ$ pillar = $150 – x$.

$$\tan 60^\circ = \frac{h}{x} \implies h = x\sqrt 3$$

$$\tan 30^\circ = \frac{h}{150 – x} \implies h\sqrt 3 = 150 – x$$

Substituting $h = x\sqrt 3$:

$$3x = 150 – x \implies 4x = 150 \implies x = 37.5$$

Height $= 37.5\sqrt 3 \approx 64.95$ m. The point is 37.5 m from one pillar and 112.5 m from the other.

Short-Answer Practice Set

SA-1

The string of a kite is 100 m long and it makes an angle of 60° with the horizontal. Find the height of the kite, assuming there is no slack in the string.

$$\sin 60^\circ = \frac{h}{100} \implies h = 100 \cdot \frac{\sqrt 3}{2} = 50\sqrt 3$$

Height of the kite is $50\sqrt 3$ m $\approx 86.6$ m.

SA-2

A kite is flying at a height of 75 m from the ground. The string makes an angle $\theta$ with the ground such that $\tan\theta = 15/8$. How long is the string?

From the right triangle, opposite = 15, adjacent = 8, hypotenuse = $\sqrt{15^2 + 8^2} = 17$. So $\sin\theta = 15/17$.

$$\sin\theta = \frac{75}{L} \implies \frac{15}{17} = \frac{75}{L} \implies L = 85\text{ m}$$

The length of the string is 85 m.

SA-3

A 5 m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4 m high. If the foot of the ladder is moved 1.6 m towards the wall, find the distance by which the top of the ladder moves up.

Initial: foot $= \sqrt{5^2 – 4^2} = 3$ m from wall, top at 4 m. New foot distance = $3 – 1.6 = 1.4$ m. New height $= \sqrt{5^2 – 1.4^2} = \sqrt{25 – 1.96} = \sqrt{23.04} = 4.8$ m.

The top moves up by $4.8 – 4 = $ 0.8 m.

SA-4

The angle of elevation of the top of a building from the foot of the tower is 30°. The angle of elevation of the top of the tower from the foot of the building is 60°. The tower is 50 m high. Find the height of the building. (Already in NCERT Q9.)

From the worked solution, height $= \dfrac{50}{3} \approx 16.67$ m.

SA-5

The horizontal distance between two towers is 60 m. The angle of depression of the first tower as observed from the top of the second tower is 30°. If the height of the second tower is 150 m, find the height of the first tower.

Let height of first tower = $h$. Vertical drop from top of tower 2 to top of tower 1 = $150 – h$.

$$\tan 30^\circ = \frac{150 – h}{60} \implies 150 – h = \frac{60}{\sqrt 3} = 20\sqrt 3$$

$$h = 150 – 20\sqrt 3 \approx 150 – 34.64 = 115.36\text{ m}$$

Height of the first tower is approximately 115.36 m.

Case-Study Style Problems

Case Study 1 — Bridge Building

An engineer is designing a suspension bridge between two cliffs across a valley. From the top of cliff A (height 80 m), the angle of depression of the foot of cliff B is 30°. The angle of elevation of the top of cliff B from the foot of cliff A is 60°.

(i) Find the horizontal distance between the cliffs.

$$\tan 30^\circ = \frac{80}{d} \implies d = 80\sqrt 3 \approx 138.56\text{ m}$$

(ii) Find the height of cliff B.

$$\tan 60^\circ = \frac{H_B}{d} \implies H_B = 80\sqrt 3 \cdot \sqrt 3 = 240\text{ m}$$

(iii) What is the height difference between the cliffs? $240 – 80 = 160$ m.

Case Study 2 — Lighthouse Vigilance

A coastguard observes from the top of a 100 m lighthouse two boats lying on the same straight line on the sea. The angles of depression of the two boats are 60° and 45°.

(i) How far is the nearer boat from the lighthouse base?

$$\tan 60^\circ = \frac{100}{d_1} \implies d_1 = \frac{100}{\sqrt 3} = \frac{100\sqrt 3}{3} \approx 57.74\text{ m}$$

(ii) How far is the farther boat?

$$\tan 45^\circ = \frac{100}{d_2} \implies d_2 = 100\text{ m}$$

(iii) Distance between boats $= d_2 – d_1 = 100 – \dfrac{100\sqrt 3}{3} = \dfrac{300 – 100\sqrt 3}{3} = \dfrac{100(3 – \sqrt 3)}{3}$ m $\approx 42.27$ m.

Case Study 3 — Drone Surveillance

A drone hovers at a height of $50\sqrt 3$ m above a level field. It records two markers A and B on the field at angles of depression of 60° and 30° respectively, on opposite sides of the drone.

(i) Distance of marker A: $\tan 60^\circ = \dfrac{50\sqrt 3}{x_A} \implies x_A = 50$ m.

(ii) Distance of marker B: $\tan 30^\circ = \dfrac{50\sqrt 3}{x_B} \implies x_B = 150$ m.

(iii) Distance between markers: $50 + 150 = 200$ m.

Multiple-Choice Questions

MCQ 1

If the angle of elevation of the top of a tower from a point at a distance of 100 m from the foot of the tower is 60°, the height of the tower is:

(A) $50$ m (B) $100\sqrt 3$ m (C) $\dfrac{100}{\sqrt 3}$ m (D) $50\sqrt 3$ m

Answer: (B). $\tan 60^\circ = \dfrac{h}{100} \implies h = 100\sqrt 3$ m.

MCQ 2

The angle of elevation of the sun, when the length of the shadow of a tree is equal to the height of the tree, is:

(A) $30^\circ$ (B) $45^\circ$ (C) $60^\circ$ (D) $90^\circ$

Answer: (B). $\tan\theta = \dfrac{h}{h} = 1 \implies \theta = 45^\circ$.

MCQ 3

If the height of a tower and the distance of the point of observation from its foot, both, are increased by 10%, then the angle of elevation of its top remains:

(A) Increased (B) Decreased (C) Same (D) Cannot be determined

Answer: (C). Both numerator and denominator scale equally, so $\tan\theta$ is unchanged.

MCQ 4

From a point on the ground, 20 m away from the foot of a vertical tower, the angle of elevation of the top of the tower is 60°. The height of the tower is:

(A) $10\sqrt 3$ m (B) $20\sqrt 3$ m (C) $30\sqrt 3$ m (D) $40$ m

Answer: (B). $h = 20\tan 60^\circ = 20\sqrt 3$ m.

MCQ 5

The angle of depression of a boat from the top of a 50 m high cliff is 30°. The distance of the boat from the foot of the cliff is:

(A) $50$ m (B) $50\sqrt 3$ m (C) $\dfrac{50}{\sqrt 3}$ m (D) $25\sqrt 3$ m

Answer: (B). $\tan 30^\circ = \dfrac{50}{d} \implies d = 50\sqrt 3$ m.

MCQ 6

If the length of the shadow of a tower is $\sqrt 3$ times the height of the tower, the angle of elevation of the sun is:

(A) $30^\circ$ (B) $45^\circ$ (C) $60^\circ$ (D) $90^\circ$

Answer: (A). $\tan\theta = \dfrac{h}{h\sqrt 3} = \dfrac{1}{\sqrt 3} \implies \theta = 30^\circ$.

MCQ 7

A pole 6 m high casts a shadow $2\sqrt 3$ m long on the ground. The sun’s elevation is:

(A) $30^\circ$ (B) $45^\circ$ (C) $60^\circ$ (D) $90^\circ$

Answer: (C). $\tan\theta = \dfrac{6}{2\sqrt 3} = \sqrt 3 \implies \theta = 60^\circ$.

MCQ 8

From the top of a 100 m high tower, the angles of depression of two cars on the same side are 30° and 45°. The distance between the cars is:

(A) $100(\sqrt 3 – 1)$ m (B) $100\sqrt 3$ m (C) $100$ m (D) $\dfrac{100}{\sqrt 3}$ m

Answer: (A). Far car at $100\sqrt 3$, near car at $100$, difference $= 100(\sqrt 3 – 1)$.

Frequently Asked Questions

Q1. What is the difference between angle of elevation and angle of depression?

The angle of elevation is measured upward from the horizontal when the object is above the observer; the angle of depression is measured downward from the horizontal when the object is below the observer. Both are measured between the line of sight and the horizontal line through the observer’s eye.

Q2. Why is the angle of depression equal to the angle of elevation in two-observer problems?

Because the horizontal at the top observer is parallel to the ground at the bottom. When the line of sight is the transversal, alternate interior angles are equal. Hence the angle of depression at A equals the angle of elevation at B.

Q3. Which trigonometric ratio should I use first?

List what is given and what is required:

Hypotenuse and vertical $\to$ use $\sin\theta$.
Hypotenuse and horizontal $\to$ use $\cos\theta$.
Vertical and horizontal $\to$ use $\tan\theta$.

For most height-and-distance problems, $\tan\theta$ is the most useful because heights and ground distances are typically the perpendicular and base of the right triangle.

Q4. The observer has some height. How do I include it?

Subtract the observer’s eye-height from the object’s height. The right triangle is formed between the observer’s eye, the foot of the object on the horizontal level of the eye, and the top of the object. The horizontal distance is the same as the ground distance.

Q5. What if the angles of elevation from two points are complementary?

Use the identity $\tan\theta \cdot \tan(90^\circ – \theta) = \tan\theta \cdot \cot\theta = 1$. Multiplying the two tan-relations eliminates the angle and lets you solve directly for the height.

Q6. How many marks are usually allotted to this chapter in the ASSEB exam?

Heights and Distances typically carries between 4 to 8 marks in the HSLC paper, often as a single 3-mark or 4-mark problem. Practising the 16 NCERT exercise questions ensures full preparation.

Q7. Are calculator-based decimal approximations acceptable?

Final answers are usually expected in surd form (e.g. $10\sqrt 3$). When the question gives $\sqrt 3 = 1.732$ or $\sqrt 2 = 1.414$, you must convert to decimal in the final step.

Common Mistakes Students Make

Confusing elevation with depression. The angle is always measured from the horizontal — never from the vertical. Drawing the horizontal dotted line first prevents this error.
Forgetting the observer’s height. If a 1.5 m boy or 1.2 m girl is mentioned, this height must be subtracted from the object’s height before applying $\tan\theta$.
Using $\sin$ when $\tan$ is correct. Heights and ground distances form the perpendicular and base — use $\tan\theta$. Use $\sin\theta$ only when the hypotenuse (rope, ladder, kite-string) is involved.
Leaving a surd in the denominator. Always rationalise: $\dfrac{30}{\sqrt 3} = 10\sqrt 3$.
Mixing up complementary angles. If two elevation angles add to $90^\circ$, set them as $\theta$ and $90^\circ – \theta$ and use $\tan(90^\circ – \theta) = \cot\theta = \dfrac{1}{\tan\theta}$.
Wrong direction of motion. When a person walks towards a building, the angle of elevation increases; when they walk away, it decreases. Match the larger angle to the closer position.
Unit errors. Always convert m/s to km/h using $\times \dfrac{18}{5}$ and km/h to m/s using $\times \dfrac{5}{18}$. Mixing these units is a common cause of lost marks.
Forgetting alternate interior angles. The angle of depression at the top equals the angle of elevation at the bottom — this lets you place the angle inside the right triangle correctly.

Real-Life Applications of Trigonometry

Trigonometry is not confined to textbook examples; it powers many real-world technologies and professions:

Surveying and Civil Engineering. Engineers use theodolites to measure angles of elevation and depression while planning roads, bridges and buildings. Heights of mountains and depths of valleys are computed exactly as in this chapter.
Astronomy. The distance to nearby stars and the height of celestial objects are measured by parallax angles — pure trigonometry.
Aviation. Pilots compute glide-path angles, descent rates and ground distances using elevation/depression relations identical to those you have just solved.
Navigation and GPS. Modern GPS uses satellite trigonometry to triangulate positions on Earth with accuracy down to a metre.
Architecture. Roof slopes, ramps for accessibility (ADA-style 1-in-12 inclines), staircases and even drainage gradients are angle-of-elevation problems.
Forestry. Foresters use a clinometer to measure the angle to the top of a tree and the horizontal distance, then compute the tree’s height — exactly Question 4 of Exercise 9.1.
Defence and Artillery. Range tables for projectiles depend on elevation angles; spotters report enemy positions by depression angles from observation posts.
Photography and Drones. Camera tilt and drone surveillance distances reduce to the same right-triangle problems studied here.

Whenever you observe something far above or below your horizontal eye-line and wonder how high or far it is, you are setting up a Chapter 9 problem. Trigonometry is the language with which engineers, scientists and surveyors converse with the physical world.

Examination Strategy

For ASSEB HSLC, the heights-and-distances question is almost always a 3-mark or 4-mark problem. To secure full marks:

Spend the first 30 seconds drawing a labelled figure with given lengths and angles. The figure itself often carries 1 mark.
Write the trigonometric equation explicitly: e.g. “In right $\triangle ABC$, $\tan 60^\circ = \dfrac{AB}{BC}$.”
Substitute standard values cleanly and rationalise.
End with a one-line conclusion: “Hence the required height is ….” This earns the final mark.

Conclusion

Some Applications of Trigonometry is one of the most enjoyable chapters in Class 10 Mathematics, because every problem corresponds to something you can imagine — a tower, a kite, a ship, a balloon. Master the four basic skills: drawing the figure, marking the angle, picking the right ratio, and substituting standard values. With the 16 NCERT exercise questions and the additional practice problems above, you have covered the full range of patterns asked in the ASSEB HSLC examination. Keep practising, and best wishes from HSLC Guru.