SOME APPLICATIONS OF TRIGONOMETRY-Exercise 9.1

Trigonometry becomes truly meaningful when it helps to answer practical questions such as “How high is that tower?” or “How far is that ship from the lighthouse?” without directly measuring the height or the distance. Chapter 9, “Some Applications of Trigonometry”, focuses on such real-life problems of heights and distances, where angles of elevation and depression are used to model situations involving buildings, trees, towers, bridges, mountains and moving objects. In this textbook exercise solutions set, every NCERT question is solved in a simple, systematic and exam-oriented manner so that students can clearly see how the diagram, right-angled triangle, and trigonometric ratios are chosen and applied step by step. Each solution highlights the identification of known and unknown quantities, selection of appropriate trigonometric ratios (sine, cosine, tangent), and neat presentation of calculations, helping students develop both conceptual understanding and speed for board examinations. The solved examples and exercises are aligned with the latest CBSE syllabus and pattern, making them suitable for Class 10 CBSE, State Boards, and other school-level examinations that follow NCERT textbooks. Students can use these solutions as a guided practice resource: first to learn the standard approach for heights-and-distances problems and then to revise quickly before tests, sample papers, pre-boards and the final board exam.

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TRIGONOMETRIC FUNCTIONS-Exercise 3.2

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Maths

TRIGONOMETRIC FUNCTIONS-Exercise 3.1

Exercise • Jan 2026

Trigonometric Functions form a crucial foundation of higher mathematics and play a vital role in physics, engineering, astronomy, and real-life proble...

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Exercise
December 13, 2025  |  By Academia Aeternum

SOME APPLICATIONS OF TRIGONOMETRY-Exercise 9.1

Maths - Exercise
Fig. 9.11
Fig. 9.11

Q1. 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° (see Fig. 9.11).

Solution:

length of the rope = 20m
Angle of Inclination of rope = 30°
Let height be (h\)

$$\begin{aligned}\sin 30^{\circ }&=\dfrac{h}{20}\\ \dfrac{1}{2}&=\dfrac{h}{20}\\ h&=\dfrac{20}{2}\\ &=10m\end{aligned}$$
Height of the pole is 10 m
Fig. 9.11.2
Fig. 9.11.2

Q2. 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.

Solution:

Angle of inclination that top of the tree makes with ground = 30°
Distance from ground = 8m
Let height of tree = \(h\)

$$\begin{align}h&=BC+AC\\ AB&=8m\\ \cos 30^{\circ }&=\dfrac{8}{AC}\\\\ \sqrt{3}&=\dfrac{8}{AC}\\\\ AC&=\dfrac{8}{\frac{\sqrt{3}}{2}}\\\\ &=\dfrac{16}{\sqrt{3}}\tag{1} \\\\ \sin 30^{\circ}&=\dfrac{BC}{AC}\\\\ BC&=AC\cdot \sin 30^{\circ }\\\\ &=\dfrac{16}{\sqrt{3}}\cdot \dfrac{1}{2}\\\\ &=\dfrac{8}{\sqrt{3}}\tag{2}\end{align}$$

Adding Equation-(1) and equation (2)

$$\begin{aligned}h&=AC+BC\\\\ &=\dfrac{16}{\sqrt{3}}+\dfrac{8}{\sqrt{3}}\\\\ &=\dfrac{24}{\sqrt{3}}\\\\ &=\dfrac{24\cdot \sqrt{3}}{\sqrt{3}x\sqrt{3}}\\\\ &=\dfrac{24\sqrt{3}}{3}\\\\ &=8\sqrt{3}\end{aligned}$$
Height of tree is \(8\sqrt{3}\) m
Fig. 9.11.3-x
Fig. 9.11.3

Q3. A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?

Solution:

Height of slide for kids below 5 year of age = 1.5 m
Angle of Inclination = 30°
Height of slide for elders = 3m
Angle of Inclination for slide of elders = 60°
length of slide for youngers

$$\begin{aligned}\sin 30^{\circ }&=\dfrac{\Lambda B}{AC}\\\\ \dfrac{1}{2}&=\dfrac{1.5}{AC}\\\\ AC&=1\cdot 5\times 2\\\\ &=3m\end{aligned}$$

Length of slides for elders

$$\begin{aligned}\sin 60^{\circ }&=\dfrac{PQ}{PR}\\\\ \dfrac{\sqrt{3}}{2}&=\dfrac{3}{PR}\\\\ PR&=\dfrac{3\times 2}{\sqrt{3}}\\\\ &=2\sqrt{3}\\\\ &=2\times 1\cdot 732\\\\ &=3\cdot 464m\end{aligned}$$
Length of the slide for kids below 5 years of age is 3m and
length of slide of elders is 3.464m
Fig. 9.11.4-x
Fig. 9.11.4

Q4. 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.

Solution:

Angle of elevation = 30°
distance of tower = 30m

$$\begin{aligned}\tan 30^{\circ }&=\dfrac{AB}{BC}\\\\ AB&=BC\cdot \tan 30^{\circ }\\\\ &=30\times \dfrac{1}{\sqrt{3}}\\\\ &=\dfrac{30\sqrt{3}}{\sqrt{3}\sqrt{3}}\\\\ &=\dfrac{30\sqrt{3}}{3}\\\\ &=10\sqrt{3}\end{aligned}$$
Height of tower is \(10\sqrt{3}\) m
Fig. 9.11.5-x
Fig. 9.11.4

Q5. 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.

Solution:

Height of flying kite = 60m
Inclination of string tied on the ground = 60°
Therefore length of string =

$$\begin{aligned}\sin 60^{\circ }&=\dfrac{AB}{AC}\\\\ \dfrac{\sqrt{3}}{2}&=\dfrac{60}{AC}\\\\ AC&=\dfrac{60\times 2}{\sqrt{3}}\\\\ &=\dfrac{60\times 2\times \sqrt{3}}{\sqrt{3}\cdot \sqrt{3}}\\\\ &=40\sqrt{3}\end{aligned}$$
Length of the string is \(= 40\sqrt{3}\)
Fig. 9.11.6-x
Fig. 9.11.6

Q6. 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.

Solution:

Height of the boy = 1.5m
Height of the building = 30m
Angle of elevation is 30° and becomes 60° as he moves toward building
Let Distance travelled by boy be \(x\) m

$$\begin{aligned}\tan 30^{\circ }&=\dfrac{AB}{BC}\\\\ AB&=30-1.5=28.5\\ \scriptsize\text{(Height }&\scriptsize\text{of Building - Height od boy)}\\\\ \dfrac{1}{\sqrt{3}}&=\dfrac{28.5}{BC}\\\\ BC&=28.5\sqrt{3}\\\\ \tan 60^{\circ }&=\dfrac{AB}{BD}\\\\ \sqrt{3}&=\dfrac{28.5}{BD}\\\\ BD&=\dfrac{28.5}{\sqrt{3}}\\\\ &=\dfrac{28.5\sqrt{3}}{\sqrt{3}\times \sqrt{3}}\\\\ &=9.5\sqrt{3}\\\\ x&=BC-BD\\\\ &=28.5\sqrt{3}-9.5\sqrt{3}\\\\ &=\sqrt{3}\left( 28.5-9.5\right) \\\\ &=\sqrt{3}\left( 19.0\right) \\\\ &=19\sqrt{3}\end{aligned}$$
Boy walked 195 m towards the building
Fig. 9.11.6-x
Fig. 9.11.6

Q7. 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.

Solution:

Let height of tower is base is \(x\)
height of the building = 20m
Agle of elevation bottom of tower 45°
Angle of elevation from top of the tower = 60°

$$\begin{aligned}\tan 45^{\circ }&=\dfrac{DC}{BC}\\\\ BC&=\dfrac{DC}{\tan 45^{\circ }}\\\\ BC&=DC\\ BC&=20m\\\\ \tan 60^{\circ }&=\dfrac{20+x}{BC}\\\\ \sqrt{3}&=\dfrac{20+x}{20}\\\\ 20\sqrt{3}&=20+x\\\\ x&=20\sqrt{3}-20\\\\ &=20\left( \sqrt{3}-1\right)\ m\end{aligned}$$
Height of the tower is \(20\left( \sqrt{3}-1\right)\ m\)
Fig. 9.11.8-x
Fig. 9.11.8

Q8. 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.

Solution:

Height of statue = 1.6m
Angle of elevation on the top of statue = 60°
Angle of elevation on the top of pedestal = 45°
Let height of Pedestal = \(x\) m

$$\begin{aligned} \tan 45^{\circ }&=\dfrac{x}{BC}\\\\ BC &=x\quad\scriptsize\left( \tan 45^{\circ }=1\right) \\\\ \tan 60^{\circ }&=\dfrac{AC}{BC}\\ \scriptsize\text{(Substituting }&\scriptsize\text{Values of AC & BC)}\\\\ \tan 60^{\circ }&=\dfrac{x+1.6}{x}\\\\ \sqrt{3}&=\dfrac{x+1.6}{x}\\\\ \sqrt{3}x-x&=1.6\\\\ x\left( \sqrt{3}-1\right) &=1.6\\\\ x&=\dfrac{1.6}{\sqrt{3}-1}\times \dfrac{\left( \sqrt{3}+1\right) }{\left( \sqrt{3}+1\right) }\\\\ &=\dfrac{1.6\cdot \sqrt{3}+1}{2}\\\\ &=0.8\left( \sqrt{3}+1\right)\ m \end{aligned}$$
Height of the pedestal = \(0.8\left( \sqrt{3}+1\right)\ m\)
Fig. 9.11.9-x
Fig. 9.11.9

Q9. 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.

Solution:

Height of tower = 50m
Angle of elevation from tower to ground = 60°
Angle of elevation from building to ground = 30°

$$\begin{aligned}\tan 60^{\circ }&=\dfrac{DC}{BC}\\\\ BC&=\dfrac{DC}{\tan 60^{\circ }}\\\\ BC&=\dfrac{50}{\sqrt{3}}\\\\ \tan 30^{\circ }&=\dfrac{AB}{BC}\\\\ \dfrac{1}{\sqrt{3}}&=\dfrac{AB}{\dfrac{50}{\sqrt{3}}}\\\\ \dfrac{1}{\sqrt{3}}&=\dfrac{\sqrt{3}\times AB}{50}\\\\ AB&=\dfrac{50}{\sqrt{3}\times \sqrt{3}}\\\\ &=\dfrac{50}{3}\\\\ &=16\dfrac{2}{3}\ m\end{aligned}$$
Height of the building is \(16\dfrac{2}{3}\) m
Fig. 9.11.10-x
Fig. 9.11.10

Q10. 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.

Solution:

Width of the road = 80 m
Angle of elevation on the poles from any point between the road is 60° and 30°
Let height of poles be \(x\),
Let distance between poles be \(y\) and \(80-y\)

$$\begin{align}\tan 30^{\circ }&=\dfrac{x}{80-y}\\\\ x&=\dfrac{1}{\sqrt{3}}\left( 80-y\right) \tag{1} \\\\ \tan 60^{\circ }&=\dfrac{x}{y}\\\\ x&=y\tan 60^{\circ }\\\\ &=y\sqrt{3}\tag{2} \\\\ \text{Equation - }\left( 1\right) &=\text{Equation - }\left( 2\right) \\\\ \dfrac{\left( 80-y\right) }{\sqrt{3}}&=\sqrt{3}y\\\\ 80-y&=3y\\\\ 80&=3y+y\\\\ 4y&=80\\\\ y&=\dfrac{80}{4}\\\\ &=20\\\\ 80-y&=80-20\\\\ &=60m\\\\ x&=\dfrac{60}{\sqrt{3}}\\\\ x&=\dfrac{60\sqrt{3}}{\sqrt{3}\cdot \sqrt{3}}\\\\ &=\dfrac{60\sqrt{3}}{3}\\\\ &=20\sqrt{3}\end{align}$$
Height of poles is \(20\sqrt{3}\) m and distance from points is 20 m and 60 m
Fig. 9.11.11-x
Fig. 9.11.11

Q11. 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 joing this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see Fig. 9.12). Find the height of the tower and the width of the canal.

Solution:

Angle of elevation from the top of tower on the other side of canal = 60°
From a point 20 m away from this angle of elevation is 30°
Let width of canal be \(x\) m

$$\begin{aligned}\tan 60^{\circ }&=\dfrac{AB}{x}\\\\ AB&=x\tan 60^{\circ }\\\\ &=x\sqrt{3}\\\\ \tan 30^{\circ }&=\dfrac{AB}{BC}\\\\ \dfrac{1}{\sqrt{3}}&=\dfrac{AB}{20+x}\\\\ AB&=\dfrac{\left( 20+x\right) }{\sqrt{3}}\\\\ x\sqrt{3}&=\left( 20+x\right) \\\\ 3x&=20+x\\\\ 3x-x&=20\\\\ 2x&=20\\\\ x&=\dfrac{20}{2}\\\\ &=10m\\\\ AB&=x\sqrt{3}\\\\ &=10\sqrt{3}\end{aligned}$$
Height of tower = \(10\sqrt{3}\) m and width of canal is \(10\) m
Fig. 9.11.12-x
Fig. 9.11.12

Q12. 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.

Solution:

Height of the building = 7m
Angle of elevation on the top of The cable tower is 60° and
angle of depression on its foot is 45°.

$$\begin{aligned}\tan 45^{\circ }&=\dfrac{AB}{BC}\\\\ 1&=\dfrac{7}{BC}\\\\ BC&=7\\ AE&=7\\\\ \tan 60^{\circ }&=\dfrac{DE}{AE}\\\\ \sqrt{3}&=\dfrac{y}{7}\\\\ y&=7\sqrt{3}\\\\ CD&=DE+CE\\\\ &=7\sqrt{3}+7\\\\ &=7\left( \sqrt{3}+1\right) \end{aligned}$$
Height of the tower is \(7\left( \sqrt{3}+1\right)\) m.
Fig. 9.11.13-x
Fig. 9.11.12

Q13. 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.

Solution:

Height of the lighthouse = 75m
Angle of depression on two ships from top is 30° and 45°
Let distance between ships is \(x\) m

$$\begin{aligned}\tan 45^{\circ }&=\dfrac{AB}{BC}\\\\ 1&=\dfrac{75}{BC}\\\\ BC&=75\\\\ \tan 30^{\circ }&=\dfrac{AB}{BD}\\\\ \scriptsize(BD&\scriptsize=BC+CD)\\\\ \dfrac{1}{\sqrt{3}}&=\dfrac{75}{75+x}\\\\ 75+x&=75\sqrt{3}\\\\ x&=75\sqrt{3}-75\\\\ &=75\left( \sqrt{3}-1\right) \end{aligned}$$
Distance between the two ships is \(75\left( \sqrt{3}-1\right) \) m
Fig. 9.13
Fig. 9.13

Q14. 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° (see Fig. 9.13). Find the distance travelled by the balloon during the interval.

Solution:

Height of girl = 1.2m
Angle of elevation 60°
After some time elevation reduces to 30°
Let the distance travelled by balloon = \(x\)
Height of balloon from girls eye is 88.2-1.2=87

$$\begin{aligned}\tan 60^{\circ }&=\dfrac{87}{AC}\\\\ \sqrt{3}&=\dfrac{87}{AC}\\\\ AC&=\dfrac{87}{\sqrt{3}}\\\\ &=\dfrac{87\times \sqrt{3}}{\sqrt{3}\cdot \sqrt{3}}\\\\ &=\dfrac{87\sqrt{3}}{3}\\\\ &=29\sqrt{3}\\\\ \tan 30^{\circ }&=\dfrac{87}{AE}\\\\ \dfrac{1}{\sqrt{3}}&=\dfrac{87}{AC+CE}\\\\ AC+CE&=87\sqrt{3}\\\\ 29\sqrt{3}+x&=87\sqrt{3}\\\\ x&=87\sqrt{3}-29\sqrt{3}\\\\ &=\sqrt{3}\left( 87-29\right) \\\\ &=\sqrt{3}\left( 58\right) \\\\ &=58\sqrt{3}\end{aligned}$$
Distance travelled by the balloon during the interval is \(58\sqrt{3}\) m
Fig. 9.11.15-x
Fig. 9.11.15

Q15. 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.

Solution:

Angle of Depression from top of tower is 30°,
after 6 s angle of depression = 60°

$$\begin{aligned}\tan 60^{\circ }&=\dfrac{AB}{BD}\\\\ AB&=BD\times\tan 60^{\circ }\\\\ &=BD\sqrt{3}\\\\ \tan 30^{\circ }&=\dfrac{AB}{BC}\\\\ AB&=BC\tan 30^{\circ }\\\\ AB&=\dfrac{BC}{\sqrt{3}}\\\\ BD\sqrt{3}&=\dfrac{BC}{\sqrt{3}}\\\\ 3BD=&BC\\\\ BC&=3BD\end{aligned}$$

Let car was initially at the distance of \(3y\), after 6s its distance is = \(y\)
Distance to travelled by the car = \(3y-y = 2y\) time taken to travel \(2y\) is = 6sec
Let speed of car = \(S\)

$$\begin{aligned}S&=\dfrac{2y}{6}\\\\ S&=\dfrac{y}{3}\\\\ T&=\dfrac{3y}{y}\\\\ &=3s\end{aligned}$$
Car will take 3s to reach the tower.

Frequently Asked Questions

To apply trigonometric ratios (sin, cos, tan) to real-life problems involving heights and distances using angles of elevation and depression.

The straight, imaginary line joining the observer’s eye to the object being viewed.

The angle formed between the horizontal line of sight and the upward line of sight when an observer looks at an object above eye level.

The angle formed between the horizontal line of sight and the downward line of sight when an observer views an object below eye level.

Because the horizontal distance and vertical height naturally form perpendicular lines, creating right triangles useful for applying trigonometric ratios.

Primarily tangent (tan ?), but sine (sin ?) and cosine (cos ?) are also used depending on known sides.

tan ? = Opposite side / Adjacent side.

When the vertical height corresponds to the opposite side and the given length is the hypotenuse.

When the horizontal distance corresponds to the adjacent side and the given length is the hypotenuse.

Only standard angles (30°, 45°, 60°) are used, whose trigonometric ratios are known.

sin 30°=½, sin 45°=v2/2, sin 60°=v3/2; cos 30°=v3/2, cos 45°=v2/2, cos 60°=½; tan 30°=1/v3, tan 45°=1, tan 60°=v3.

Draw a clear, labeled diagram converting the scenario into a right triangle.

It helps identify the unknown side, the angle given, and the correct trigonometric ratio to use.

The imaginary line parallel to the ground passing through the observer’s eye.

The person, point, or object from which sight or measurement is taken.

Towers, poles, trees, buildings, mountains, ships, airplanes, balloons, and bridges.

A distance that cannot be measured directly, requiring trigonometric methods.

Yes, by using angles of elevation or depression from a known point.

Two-point observation method, resulting in two different right triangles.

Use tan ? = height/distance ? height = distance × tan ?.

distance = height / tan ?.

Viewing the top of a tower or kite from ground level.

Seeing a car from a lighthouse balcony.

Observing the same object from two different positions, yielding two different elevation angles.

As the observer moves closer, the angle of elevation increases.

Right-triangle geometry and trigonometric ratios.

Finding height, distance, width, altitude, or length using given angles of elevation or depression.

Sometimes, when two sides are known or trigonometric ratios are insufficient.

tan 45° = 1, simplifying height = distance.

tan 30° = 1/v3, often appearing in height problems requiring rationalization.

The object is vertically above the observer—distance is zero (mostly theoretical).

Opposite is the vertical side from angle; adjacent is the horizontal side.

NCERT restricts problems to standard angles with known ratios.

Theodolite, used by surveyors to measure angles.

To estimate altitude and distance from the ground.

For determining distance of ships, lighthouses, and ports.

The syllabus focuses only on elementary applications—heights and distances.

Single angle of elevation with either height or distance known.

When two right triangles share a common vertical line or horizontal line.

A real-life scenario with one or two angles and one unknown distance, requiring diagram + calculation.

Convert angle of depression to angle of elevation at the lower point (alternate interior angles).

Because they form alternate interior angles with horizontal lines.

Use tan ? = shadow length / height or vice versa.

Yes, especially when forming equations from the triangle.

To provide practical, realistic measurements.

Yes, depending on the triangle setup and chosen trigonometric ratio.

Height represents perpendicular distance from ground to the object’s top.

Yes, by modelling below-ground or underwater observations.

It strengthens understanding of angles and their variation.

Neat diagram, correct ratio selection, accurate steps, and clear final answer with units.

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