---
title: "NCERT Solutions for Class 10 Maths Chapter 9 Exercise 9.1"
url: https://www.swavid.com/maths/class/10/chapter/some-applications-of-trigonometry/ncert-solutions/exercise-9-1
dateModified: 2026-10-07T15:53:18+00:00
---

# NCERT Solutions for Class 10 Maths Chapter 9 Exercise 9.1

Chapter 9: Some Applications of Trigonometry. Every question from Exercise 9.1, with full working and the final answer.

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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.

## Related pages

- [All Chapter 9 NCERT solutions](https://www.swavid.com/maths/class/10/chapter/some-applications-of-trigonometry/ncert-solutions)

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