---
title: "NCERT Solutions Class 10 Maths Some Applications of Trigonometry"
url: https://www.swavid.com/maths/class/10/chapter/some-applications-of-trigonometry/ncert-solutions
dateModified: 2026-10-07T15:53:18+00:00
---

# NCERT Solutions Class 10 Maths Some Applications of Trigonometry

This chapter's questions cover various applications of trigonometry, specifically focusing on calculating heights and distances using angles of elevation and depression.

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## EXERCISE 9.1

### Question 1

*3 marks · Short answer*

A circus artist is climbing a $20\text{ m}$ 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 $30^\circ$ (see Fig. 9.11).

**Solution**

1. Let AB represent the vertical pole and AC represent the $20\text{ m}$ long rope such that $\angle ACB = 30^\circ$.
2. In right $\triangle ABC$, $\frac{\text{AB}}{\text{AC}} = \sin 30^\circ$.
3. Substitute the values to get $\frac{\text{AB}}{20} = \frac{1}{2}$.
4. Therefore, $\text{AB} = \frac{20}{2} = 10\text{ m}$.
5. Hence, the height of the pole is $10\text{ m}$.

**Answer:** $10\text{ m}$

> Common mistake: Using cosine or tangent instead of sine ratio.

### Question 2

*3 marks · Short answer*

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

**Solution**

1. Let $AC$ be the unbroken part of the tree and $AB$ be the broken part which bends and touches the ground at point $B$, making an angle of $30^\circ$ with the ground.
2. The distance between the foot of the tree and the point where the top touches the ground is given as $CB = 8\text{ m}$.
3. In right-angled triangle $\triangle ABC$, $\cos 30^\circ = \frac{CB}{AB}$ and $\tan 30^\circ = \frac{AC}{CB}$.
4. $\frac{\sqrt{3}}{2} = \frac{8}{AB}$, which gives $AB = \frac{16}{\sqrt{3}}\text{ m}$.
5. $\frac{1}{\sqrt{3}} = \frac{AC}{8}$, which gives $AC = \frac{8}{\sqrt{3}}\text{ m}$.
6. The total height of the tree is $AC + AB = \frac{8}{\sqrt{3}} + \frac{16}{\sqrt{3}} = \frac{24}{\sqrt{3}} = 8\sqrt{3}\text{ m}$.

**Answer:** $8\sqrt{3}\text{ m}$

> Common mistake: Adding only the length of the broken part or only the remaining part instead of their sum to find the total height of the tree.

### Question 3

*3 marks · Short answer*

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 $1.5\text{ m}$, and is inclined at an angle of $30^\circ$ to the ground, whereas for elder children, she wants to have a steep slide at a height of $3\text{ m}$, and inclined at an angle of $60^\circ$ to the ground. What should be the length of the slide in each case?

**Solution**

1. For children below 5 years, let the length of the slide be $l_1$, height be $1.5\text{ m}$ and angle be $30^\circ$.
2. In the right triangle for the first slide, $\frac{1.5}{l_1} = \sin 30^\circ = \frac{1}{2}$, giving $l_1 = 3\text{ m}$.
3. For elder children, let the length of the slide be $l_2$, height be $3\text{ m}$ and angle be $60^\circ$.
4. In the right triangle for the second slide, $\frac{3}{l_2} = \sin 60^\circ = \frac{\sqrt{3}}{2}$, giving $l_2 = \frac{6}{\sqrt{3}} = 2\sqrt{3}\text{ m}$.
5. Hence, the lengths of the slides are $3\text{ m}$ and $2\sqrt{3}\text{ m}$ respectively.

**Answer:** $3\text{ m}$ and $2\sqrt{3}\text{ m}$

> Common mistake: Using cosine or tangent instead of sine for slide length and height.

### Question 4

*3 marks · Short answer*

The angle of elevation of the top of a tower from a point on the ground, which is $30\text{ m}$ away from the foot of the tower, is $30^\circ$. Find the height of the tower.

**Solution**

1. Let AB represent the tower and C be the point on the ground at a distance of $30\text{ m}$ from the foot of the tower.
2. Here, $\text{BC} = 30\text{ m}$ and $\angle ACB = 30^\circ$.
3. In right $\triangle ABC$, $\frac{\text{AB}}{\text{BC}} = \tan 30^\circ$.
4. Substitute the values to get $\frac{\text{AB}}{30} = \frac{1}{\sqrt{3}}$, which gives $\text{AB} = \frac{30}{\sqrt{3}} = 10\sqrt{3}\text{ m}$.
5. Hence, the height of the tower is $10\sqrt{3}\text{ m}$.

**Answer:** $10\sqrt{3}\text{ m}$

> Common mistake: Rationalizing incorrectly or using sine instead of tangent.

### Question 5

*3 marks · Short answer*

A kite is flying at a height of $60\text{ m}$ 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 $60^\circ$. Find the length of the string, assuming that there is no slack in the string.

**Solution**

1. Let the height of the kite above the ground be $AB = 60\text{ m}$ and let $AC$ be the length of the string.
2. In right-angled triangle $ABC$, the angle of inclination with the ground is $\angle ACB = 60^\circ$.
3. Using the trigonometric ratio involving perpendicular and hypotenuse, $\sin 60^\circ = \frac{AB}{AC}$.
4. Substituting the values, $\frac{\sqrt{3}}{2} = \frac{60}{AC}$ which gives $AC = \frac{120}{\sqrt{3}}$.
5. Rationalising the denominator, $AC = \frac{120\sqrt{3}}{3} = 40\sqrt{3}\text{ m}$.

**Answer:** $40\sqrt{3}\text{ m}$

> Common mistake: Using cosine or tangent instead of sine for the relation between perpendicular and hypotenuse.

### Question 6

*3 marks · Short answer*

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

**Solution**

1. Let AB be the $30\text{ m}$ tall building and the boy initially stand at point C, then walk to point D.
2. The height of the boy is $1.5\text{ m}$, so the height of the building from his eye level is $\text{AE} = 30 - 1.5 = 28.5\text{ m}$.
3. In right $\triangle AEF$, $\tan 60^\circ = \frac{\text{AE}}{\text{EF}}$, giving $\text{EF} = \frac{28.5}{\sqrt{3}}\text{ m}$.
4. In right $\triangle AEC$, $\tan 30^\circ = \frac{\text{AE}}{\text{EC}}$, giving $\text{EC} = 28.5\sqrt{3}\text{ m}$.
5. The distance he walked is $\text{FC} = \text{EC} - \text{EF} = 28.5\sqrt{3} - \frac{28.5}{\sqrt{3}} = 28.5 \times \frac{2}{\sqrt{3}} = 19\sqrt{3}\text{ m}$.

**Answer:** $19\sqrt{3}\text{ m}$

> Common mistake: Forgetting to subtract the height of the boy from the total height of the building.

### Question 7

*3 marks · Short answer*

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 $20\text{ m}$ high building are $45^\circ$ and $60^\circ$ respectively. Find the height of the tower.

**Solution**

1. Let $DC$ be the $20\text{ m}$ high building and $BC$ be the transmission tower of height $h\text{ m}$ fixed at the top.
2. Let $A$ be a point on the ground at a distance $x$ from the foot of the building $D$.
3. In right-angled triangle $ADC$, $\tan 45^\circ = \frac{DC}{AD}$ which gives $1 = \frac{20}{x}$, so $x = 20\text{ m}$.
4. In right-angled triangle $ADB$, $\tan 60^\circ = \frac{DB}{AD} = \frac{20 + h}{20}$.
5. Substituting $\tan 60^\circ = \sqrt{3}$, we get $\sqrt{3} = \frac{20 + h}{20}$, which gives $h = 20(\sqrt{3} - 1)\text{ m}$.

**Answer:** $20(\sqrt{3} - 1)\text{ m}$

> Common mistake: Taking the base angle of $60^\circ$ for the building instead of the entire tower.

### Question 8

*3 marks · Short answer*

A statue, $1.6\text{ m}$ 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 $60^\circ$ and from the same point the angle of elevation of the top of the pedestal is $45^\circ$. Find the height of the pedestal.

**Solution**

1. Let $AB$ be the pedestal of height $h\text{ m}$ and $BC$ be the $1.6\text{ m}$ tall statue on top of it.
2. Let $D$ be a point on the ground at a distance $x$ from the foot of the pedestal.
3. In right-angled triangle $ABD$, $\tan 45^\circ = \frac{AB}{AD}$ which gives $1 = \frac{h}{x}$, so $x = h$.
4. In right-angled triangle $ACD$, $\tan 60^\circ = \frac{AC}{AD} = \frac{h + 1.6}{x}$.
5. Substituting $x = h$ and $\tan 60^\circ = \sqrt{3}$, we get $\sqrt{3}h = h + 1.6$, leading to $h = \frac{1.6}{\sqrt{3} - 1} = 0.8(\sqrt{3} + 1)\text{ m}$.

**Answer:** $0.8(\sqrt{3} + 1)\text{ m}$

> Common mistake: Forgetting to rationalise the denominator or combining the statue and pedestal heights incorrectly.

### Question 9

*3 marks · Short answer*

The angle of elevation of the top of a building from the foot of the tower is $30^\circ$ and the angle of elevation of the top of the tower from the foot of the building is $60^\circ$. If the tower is $50\text{ m}$ high, find the height of the building.

**Solution**

1. Let $AB$ be the building of height $h\text{ m}$ and $CD$ be the tower of height $50\text{ m}$. Let the distance between the building and the tower be $x\text{ m}$.
2. In the right-angled triangle $DCB$, the angle of elevation of the top of the tower from the foot of the building is $60^\circ$, so $\tan 60^\circ = \frac{CD}{BC}$ which gives $\sqrt{3} = \frac{50}{x}$, and $x = \frac{50}{\sqrt{3}}\text{ m}$.
3. In the right-angled triangle $ABC$, the angle of elevation of the top of the building from the foot of the tower is $30^\circ$, so $\tan 30^\circ = \frac{AB}{BC}$ which gives $\frac{1}{\sqrt{3}} = \frac{h}{x}$.
4. Substitute $x = \frac{50}{\sqrt{3}}$ into the equation to get $\frac{1}{\sqrt{3}} = \frac{h}{\frac{50}{\sqrt{3}}}$.
5. Solving for $h$ gives $h = \frac{50}{3}\text{ m} = 16\frac{2}{3}\text{ m}$.

**Answer:** $16\frac{2}{3}\text{ m}$

> Common mistake: Mixing up the angles for the building and the tower.

### Question 10

*3 marks · Short answer*

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

**Solution**

1. Let $AB$ and $CD$ be two poles of equal height $h\text{ m}$ standing on either side of a road of width $80\text{ m}$. Let $P$ be a point on the road such that $BP = x\text{ m}$, then $DP = (80 - x)\text{ m}$.
2. In the right-angled triangle $ABP$, we have $\tan 60^\circ = \frac{AB}{BP}$ which gives $\sqrt{3} = \frac{h}{x}$, so $h = x\sqrt{3}$.
3. In the right-angled triangle $CDP$, we have $\tan 30^\circ = \frac{CD}{DP}$ which gives $\frac{1}{\sqrt{3}} = \frac{h}{80 - x}$, so $h = \frac{80 - x}{\sqrt{3}}$.
4. Equating the two expressions for $h$, we get $x\sqrt{3} = \frac{80 - x}{\sqrt{3}}$, which simplifies to $3x = 80 - x$, giving $4x = 80$ or $x = 20\text{ m}$.
5. Thus, the distance of the point from the first pole is $20\text{ m}$, from the second pole is $80 - 20 = 60\text{ m}$, and the height of the poles is $h = 20\sqrt{3}\text{ m}$.

**Answer:** Height $= 20\sqrt{3}\text{ m}$, Distances $= 20\text{ m}$ and $60\text{ m}$

> Common mistake: Taking $DP = x + 80$ instead of $80 - x$.

### Question 11

*3 marks · Short answer*

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 $60^\circ$. From another point $20\text{ m}$ 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 $30^\circ$ (see Fig. 9.12). Find the height of the tower and the width of the canal.

**Solution**

1. Let $AB$ be the TV tower of height $h$ and let the width of the canal be $BC = x$. Let $D$ be another point $20\text{ m}$ away from $C$ on the line $DC$, so $DC = 20\text{ m}$.
2. In the right-angled triangle $ABC$, we have $\tan 60^\circ = \frac{AB}{BC}$ which gives $\sqrt{3} = \frac{h}{x}$, or $h = x\sqrt{3}$.
3. In the right-angled triangle $ABD$, we have $\tan 30^\circ = \frac{AB}{BD}$ which gives $\frac{1}{\sqrt{3}} = \frac{h}{20 + x}$, or $h = \frac{20 + x}{\sqrt{3}}$.
4. Equating the two values for $h$, we get $x\sqrt{3} = \frac{20 + x}{\sqrt{3}}$, which simplifies to $3x = 20 + x$, giving $2x = 20$ or $x = 10\text{ m}$.
5. Substituting $x = 10$ gives $h = 10\sqrt{3}\text{ m}$, so the height of the tower is $10\sqrt{3}\text{ m}$ and the width of the canal is $10\text{ m}$.

**Answer:** Height $= 10\sqrt{3}\text{ m}$, Width $= 10\text{ m}$

> Common mistake: Using $20$ as the distance $BC$ instead of $DC$.

### Question 12

*3 marks · Short answer*

From the top of a $7\text{ m}$ high building, the angle of elevation of the top of a cable tower is $60^\circ$ and the angle of depression of its foot is $45^\circ$. Determine the height of the tower.

**Solution**

1. Let $AB$ be the $7\text{ m}$ high building and $CD$ be the cable tower. Let $E$ be a point on the tower such that $AE \parallel BD$, which makes $BD = AE$ and $AB = ED = 7\text{ m}$.
2. In the right-angled triangle $ABD$, we have $\tan 45^\circ = \frac{AB}{BD}$ which gives $1 = \frac{7}{BD}$, so $BD = 7\text{ m}$, which means $AE = 7\text{ m}$.
3. In the right-angled triangle $ACE$, we have $\tan 60^\circ = \frac{CE}{AE}$ which gives $\sqrt{3} = \frac{CE}{7}$, so $CE = 7\sqrt{3}\text{ m}$.
4. The total height of the tower is $CD = CE + ED = (7\sqrt{3} + 7)\text{ m} = 7(\sqrt{3} + 1)\text{ m}$.
5. Therefore, the height of the tower is $7(\sqrt{3} + 1)\text{ m}$.

**Answer:** $7(\sqrt{3} + 1)\text{ m}$

> Common mistake: Omitting the height of the building $ED$ from the total height of the tower.

### Question 13

*3 marks · Short answer*

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

**Solution**

1. Let AB be the lighthouse of height $75\text{ m}$, and let C and D be the positions of the two ships such that $\angle ACB = 45^\circ$ and $\angle ADB = 30^\circ$.
2. In right $\triangle ABC$, $\frac{AB}{BC} = \tan 45^\circ = 1$, which gives $BC = AB = 75\text{ m}$.
3. In right $\triangle ABD$, $\frac{AB}{BD} = \tan 30^\circ = \frac{1}{\sqrt{3}}$, which gives $BD = AB \times \sqrt{3} = 75\sqrt{3}\text{ m}$.
4. The distance between the two ships is $CD = BD - BC = 75\sqrt{3} - 75 = 75(\sqrt{3} - 1)\text{ m}$.

**Answer:** $75(\sqrt{3} - 1)\text{ m}$

> Common mistake: Subtracting the angles instead of subtracting the distances from the right triangles.

### Question 14

*3 marks · Short answer*

A $1.2\text{ m}$ tall girl spots a balloon moving with the wind in a horizontal line at a height of $88.2\text{ m}$ from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is $60^\circ$. After some time, the angle of elevation reduces to $30^\circ$ (see Fig. 9.13). Find the distance travelled by the balloon during the interval.

**Solution**

1. Let $EF$ be the girl of height $1.2\text{ m}$ and $CD$ be the initial position of the balloon at a height of $88.2\text{ m}$ from the ground.
2. The height of the balloon from the eyes of the girl is $AB = 88.2 - 1.2 = 87\text{ m}$.
3. In right-angled triangle $ABC$ with angle of elevation $60^\circ$, $\tan 60^\circ = \frac{AB}{AC}$ which gives $\sqrt{3} = \frac{87}{AC}$, so $AC = \frac{87}{\sqrt{3}}$.
4. When the balloon moves to $P$, its height from the eyes of the girl remains $PQ = 87\text{ m}$, and the angle of elevation becomes $30^\circ$ in triangle $APQ$.
5. Therefore, $\tan 30^\circ = \frac{PQ}{AQ}$, which gives $\frac{1}{\sqrt{3}} = \frac{87}{AQ}$, so $AQ = 87\sqrt{3}\text{ m}$.
6. The distance traveled by the balloon is $CP = AQ - AC = 87\sqrt{3} - \frac{87}{\sqrt{3}} = \frac{87(3 - 1)}{\sqrt{3}} = 58\sqrt{3}\text{ m}$.

**Answer:** $58\sqrt{3}\text{ m}$

> Common mistake: Failing to subtract the girl's height of $1.2\text{ m}$ from the total height of the balloon.

### Question 15

*3 marks · Short answer*

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 $30^\circ$, 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 $60^\circ$. Find the time taken by the car to reach the foot of the tower from this point.

**Solution**

1. Let AB be the tower of height $h$ and let the speed of the car be $v$.
2. Let the initial position of the car be C at time $t = 0$ with $\angle ACB = 60^\circ$, and let its position after $6\text{ seconds}$ be D with $\angle ADB = 30^\circ$.
3. In right $\triangle ABC$, $\frac{AB}{BC} = \tan 60^\circ = \sqrt{3}$, which gives $BC = \frac{h}{\sqrt{3}}$.
4. In right $\triangle ABD$, $\frac{AB}{BD} = \tan 30^\circ = \frac{1}{\sqrt{3}}$, which gives $BD = h\sqrt{3}$.
5. The distance covered in $6\text{ seconds}$ is $CD = BD - BC = h\sqrt{3} - \frac{h}{\sqrt{3}} = \frac{2h}{\sqrt{3}}$.
6. Since the car travels $\frac{2h}{\sqrt{3}}$ in $6\text{ seconds}$, the remaining distance $BC = \frac{h}{\sqrt{3}}$ (which is half of $CD$) will take half the time, i.e., $3\text{ seconds}$.

**Answer:** $3\text{ seconds}$

> Common mistake: Assuming distance is directly proportional to angle instead of cotangent or tangent ratios.

## Frequently asked questions

### How many exercises and questions are there in Class 10 Maths Chapter 9 for the 2026-27 session?

This chapter in the NCERT textbook contains one main exercise, Exercise 9.1, which has a total of 15 questions. SwaVid provides complete step-by-step solutions for all these questions on this page only.

### What topics do the questions cover in Class 10 Maths Chapter 9?

The questions cover applications of trigonometry, right triangles, and heights and distances. Specific concepts include finding the height of a tower using the tangent ratio, horizontal motion of objects, and solving simultaneous equations in triangles where the angle of elevation changes as the observer moves.

### Which are the hardest question types in this chapter and how should students approach them?

The most challenging questions involve simultaneous equations in triangles or changing angles of elevation as an observer moves closer to or further from an object. To approach them, you should first draw a clear geometric diagram, label all given angles and distances, and set up two separate right triangles using trigonometric ratios like $\tan \theta$.

### How can students write answers for full marks in Class 10 board exams for this chapter?

To secure full marks, always start by drawing a neat and properly labeled figure representing the given word problem. Clearly state the trigonometric ratios used and write intermediate steps with proper units before arriving at the final answer.

### Is a free PDF available for Class 10 Maths Chapter 9 Some Applications of Trigonometry?

Yes, SwaVid offers a free PDF containing detailed solutions for this chapter. You can easily access and download these step-by-step solutions directly on this page only.

## Related pages

- [Exercise 9.1 solutions](https://www.swavid.com/maths/class/10/chapter/some-applications-of-trigonometry/ncert-solutions/exercise-9-1)
- [Some Applications of Trigonometry: CBSE previous year questions](https://www.swavid.com/cbse/class-10/maths/pyq/some-applications-of-trigonometry)
- [Class 10 Maths chapters](https://www.swavid.com/maths/class/10)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student