# NCERT Solutions For Class 10 Maths Chapter 9 Some Applications of Trigonometry

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**NCERT Solutions for Class 10 Maths Chapter 9 โ Some Applications of Trigonometry** provides comprehensive solutions for all the questions in the **NCERT** textbook. To excel in the board examinations, NCERT Solutions will increase the level of confidence among the students, as the concepts are clearly explained and structured. The solutions are prepared and reviewed by our subject matter experts and they are revised according to the latest NCERT syllabus and guidelines of the **CBSE board**.

It covers all the chapters and provides chapter wise solutions. These **NCERT Solutions** are helpful for the students to clarify their doubts and provide a strong foundation for every concept. With the help of NCERT Solutions for Class 10, every student should be capable of solving the complex problem in each exercise.Chapter 9 Some Applications of Trigonometry

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## Exercise 9.1 Page No: 203

**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 30ยฐ. (see fig. 9.11)**

**Solution:**

Length of the rope is 20 m and angle made by the rope with the ground level is 30ยฐ.

**Given: **AC = 20 m and angle C = 30ยฐ

**To Find**: Height of the pole

Let AB be the vertical pole

In right ฮABC, using sine formula

sin 30ยฐ = AB/AC

^{Using value of sin 30 degrees is ยฝ, we have}

^{1/2 = AB/20}

^{AB = 20/2}

^{AB = 10}

Therefore, the height of the pole is 10 m.

**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 30ยฐ 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:**

Using given instructions, draw a figure. Let AC be the broken part of the tree. Angle C = 30ยฐ

BC = 8 m

To Find: Height of the tree, which is AB

From figure: Total height of the tree is the sum of AB and AC i.e. AB+AC

In right ฮABC,

Using Cosine and tangent angles,

cos 30ยฐ = BC/AC

We know that, cos 30ยฐ = โ3/2

โ3/2 = 8/AC

AC = 16/โ3 โฆ(1)

Also,

tan 30ยฐ = AB/BC

1/โ3 = AB/8

AB = 8/โ3 โฆ.(2)

Therefore, total height of the tree = AB + AC = 16/โ3 + 8/โ3 = 24/โ3 = 8โ3 m.

**3. 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 30ยฐ 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 60ยฐ to the ground. What should be the length of the slide in each case?**

**Solution**:

As per contractorโs plan,

Let, ABC is the slide inclined at 30ยฐ with length AC and PQR is the slide inclined at

60ยฐ with length PR.

To Find: AC and PR

In right ฮABC,

sin 30ยฐ = AB/AC

1/2 = 1.5/AC

AC = 3

Also,

In right ฮPQR,

sin 60ยฐ = PQ/PR

โ โ3/2 = 3/PR

โ PR = 2โ3

Hence, length of the slide for below 5 = 3 m and

Length of the slide for elders children = 2โ3 m

**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 30ยฐ. Find the height of the tower.**

**Solution:**

Let AB be the height of the tower and C is the point elevation which is 30 m away from the foot of the tower.

To Find: AB (height of the tower)

In right ABC

tan 30ยฐ = AB/BC

1/โ3 = AB/30

โ AB = 10โ3

Thus, the height of the tower is 10โ3 m.

**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 60ยฐ. Find the length of the string, assuming that there is no slack in the string.**

**Solution:**

Draw a figure, based on given instruction,

Let BC = Height of the kite from the ground, BC = 60 m

AC = Inclined length of the string from the ground and

A is the point where string of the kite is tied.

To Find: Length of the string from the ground i.e. the value of AC

From the above figure,

sin 60ยฐ = BC/AC

โ โ3/2 = 60/AC

โ AC = 40โ3 m

Thus, the length of the string from the ground is 40โ3 m.

**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 30ยฐ to 60ยฐ as he walks towards the building. Find the distance he walked towards the building.**

**Solution:**

Let the boy initially stand at point Y with inclination 30ยฐ and then he approaches the building to the point X with inclination 60ยฐ.

To Find: The distance boy walked towards the building i.e. XY

From figure,

XY = CD.

Height of the building = AZ = 30 m.

AB = AZ โ BZ = 30 โ 1.5 = 28.5

Measure of AB is 28.5 m

In right ฮABD,

tan 30ยฐ = AB/BD

1/โ3 = 28.5/BD

BD = 28.5โ3 m

Again,

In right ฮABC,

tan 60ยฐ = AB/BC

โ3 = 28.5/BC

BC = 28.5/โ3 = 28.5โ3/3

Therefore, the length of BC is 28.5โ3/3 m.

XY = CD = BD โ BC = (28.5โ3-28.5โ3/3) = 28.5โ3(1-1/3) = 28.5โ3 ร 2/3 = 57/โ3 = 19โ3 m.

Thus, the distance boy walked towards the building is 19โ3 m.

**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 45ยฐ and 60ยฐ respectively. Find the height of the tower.**

**Solution:**

Let BC be the 20 m high building.

D is the point on the ground from where the elevation is taken.

Height of transmission tower = AB = AC โ BC

To Find: AB, Height of the tower

From figure, In right ฮBCD,

tan 45ยฐ = BC/CD

1 = 20/CD

CD = 20

Again,

In right ฮACD,

tan 60ยฐ = AC/CD

โ3 = AC/20

AC = 20โ3

Now, AB = AC โ BC = (20โ3-20) = 20(โ3-1)

Height of transmission tower = 20(โ3 โ 1) m.

**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 60ยฐ and from the same point the angle of elevation of the top of the pedestal is 45ยฐ. Find the height of the pedestal.**

**Solution:**

Let AB be the height of statue.

D is the point on the ground from where the elevation is taken.

To Find: Height of pedestal = BC = AC-AB

From figure,

In right triangle BCD,

tan 45ยฐ = BC/CD

^{1 = BC/CD}

^{BC = CD โฆ..(1)}

Again,

In right ฮACD,

tan 60ยฐ = AC/AD

^{โ3 = ( AB+BC)/CD}

^{โ3CD = 1.6 + BC}

^{โ3BC = 1.6 + BC (using equation (1)}

^{โ3BC โ BC = 1.6}

^{BC(โ3-1) = 1.6}

^{BC = 1.6/(โ3-1) m}

^{BC = 0.8(โ3+1)}

Thus, the height of the pedestal is 0.8(โ3+1) m.

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

**Solution:**

Let CD be the height of the tower. AB be the height of the building. BC be the distance between the foot of the building and the tower. Elevation is 30 degree and 60 degree from the tower and the building respectively.

In right ฮBCD,

tan 60ยฐ = CD/BC

^{โ3 = 50/BC}

^{BC = 50/โ3 โฆ(1)}

Again,

In right ฮABC,

tan 30ยฐ = AB/BC

^{โ 1/โ3 = AB/BC}

Use result obtained in equation (1)

^{AB = 50/3}

Thus, the height of the building is 50/3 m.

**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 60ยฐ and 30ยฐ, respectively. Find the height of the poles and the distances of the point from the poles.**

**Solution:**

Let AB and CD be the poles of equal height.

O is the point between them from where the height of elevation taken. BD is the distance between the poles.

As per above figure, AB = CD,

OB + OD = 80 m

Now,

In right ฮCDO,

tan 30ยฐ = CD/OD

1/โ3 = CD/OD

CD = OD/โ3 โฆ (1)

Again,

In right ฮABO,

tan 60ยฐ = AB/OB

โ3 = AB/(80-OD)

AB = โ3(80-OD)

AB = CD (Given)

โ3(80-OD) = OD/โ3 (Using equation (1))

3(80-OD) = OD

240 โ 3 OD = OD

4 OD = 240

OD = 60

Putting the value of OD in equation (1)

CD = OD/โ3

CD = 60/โ3

CD = 20โ3 m

Also,

OB + OD = 80 m

โ OB = (80-60) m = 20 m

Thus, the height of the poles are 20โ3 m and distance from the point of elevation are 20 m and

60 m respectively.

**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 60ยฐ. 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 30ยฐ (see Fig. 9.12). Find the height of the tower and the width of the canal.**

Solution: Given, AB is the height of the tower.

DC = 20 m (given)

As per given diagram, In right ฮABD,

tan 30ยฐ = AB/BD

1/โ3 = AB/(20+BC)

AB = (20+BC)/โ3 โฆ (i)

Again,

In right ฮABC,

tan 60ยฐ = AB/BC

โ3 = AB/BC

AB = โ3 BC โฆ (ii)

From equation (i) and (ii)

โ3 BC = (20+BC)/โ3

3 BC = 20 + BC

2 BC = 20

BC = 10

Putting the value of BC in equation (ii)

AB = 10โ3

This implies, the height of the tower is 10โ3 m and the width of the canal is 10 m.

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

**Solution**:

Let AB be the building of height 7 m and EC be the height of the tower.

A is the point from where elevation of tower is 60ยฐ and the angle of depression of its foot is 45ยฐ.

EC = DE + CD

Also, CD = AB = 7 m. and BC = AD

To Find: EC = Height of the tower

Design a figure based on given instructions:

In right ฮABC,

tan 45ยฐ = AB/BC

1= 7/BC

BC = 7

Since BC = AD

So AD = 7

Again, from right triangle ADE,

tan 60ยฐ = DE/AD

โ3 = DE/7

โ DE = 7โ3 m

Now: EC = DE + CD

= (7โ3 + 7) = 7(โ3+1)

Therefore, Height of the tower is 7(โ3+1) m. Answer!

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

**Solution:**

Let AB be the lighthouse of height 75 m. Let C and D be the positions of the ships.

30ยฐ and 45ยฐ are the angles of depression from the lighthouse.

Draw a figure based on given instructions:

To Find: CD = distance between two ships

Step 1: From right triangle ABC,

tan 45ยฐ = AB/BC

1= 75/BC

BC = 75 m

Step 2: Form right triangle ABD,

tan 30ยฐ = AB/BD

1/โ3 = 75/BD

BD = 75โ3

Step 3: To find measure of CD, use results obtained in step 1 and step 2.

CD = BD โ BC = (75โ3 โ 75) = 75(โ3-1)

The distance between the two ships is 75(โ3-1) m. Answer!

**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 60ยฐ. After some time, the angle of elevation reduces to 30ยฐ (see Fig. 9.13). Find the distance travelled by the balloon during the interval.**

**Solution:**

Let the initial position of the balloon be A and final position be B.

Height of balloon above the girl height = 88.2 m โ 1.2 m = 87 m.

To Find: Distance travelled by the balloon = DE = CE โ CD

Let us redesign the given figure as per our convenient

Step 1: In right ฮBEC,

tan 30ยฐ = BE/CE

1/โ3= 87/CE

CE = 87โ3

Step 2:

In right ฮADC,

tan 60ยฐ = AD/CD

โ3= 87/CD

CD = 87/โ3 = 29โ3

Step 3:

DE = CE โ CD = (87โ3 โ 29โ3) = 29โ3(3 โ 1) = 58โ3

Distance travelled by the balloon = 58โ3 m.

**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 30ยฐ, 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 60ยฐ. Find the time taken by the car to reach the foot of the tower from this point.**

**Solution:**

Let AB be the tower.

D is the initial and C is the final position of the car respectively.

Since man is standing at the top of the tower so, Angles of depression are measured from A.

BC is the distance from the foot of the tower to the car.

Step 1: In right ฮABC,

tan 60ยฐ = AB/BC

โ3 = AB/BC

BC = AB/โ3

AB = โ3 BC

Step 2:

In right ฮABD,

tan 30ยฐ = AB/BD

1/โ3 = AB/BD

AB = BD/โ3

Step 3: Form step 1 and Step 2, we have

โ3 BC = BD/โ3 (Since LHS are same, so RHS are also same)

3 BC = BD

3 BC = BC + CD

2BC = CD

or BC = CD/2

Here, distance of BC is half of CD. Thus, the time taken is also half.

Time taken by car to travel distance CD = 6 sec. Time taken by car to travel BC = 6/2 = 3 sec.

**16. The angles of elevation of the top of a tower from two points at a distance 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.**

**Solution:**

Let AB be the tower. C and D be the two points with distance 4 m and 9 m from the base respectively. As per question,

In right ฮABC,

tan x = AB/BC

tan x = AB/4

AB = 4 tan x โฆ (i)

Again, from right ฮABD,

tan (90ยฐ-x) = AB/BD

cot x = AB/9

AB = 9 cot x โฆ (ii)

Multiplying equation (i) and (ii)

AB^{2} = 9 cot x ร 4 tan x

โ AB^{2} = 36 (because cot x = 1/tan x

โ AB = ยฑ 6

Since height cannot be negative. Therefore, the height of the tower is 6 m.

Hence Proved.

## NCERT Solutions for Class 10 Maths Chapter 9 โ Some Applications of Trigonometry

For **Class 10 CBSE Maths** paper, out of the 80 marks, 12 marks are assigned from the chapter โ Trigonometryโ. You can expect 1 question from this chapter. The paper consists of 4 parts and each carries different marks. The questions have been assigned with 1 mark, two marks, 3 marks and 4 marks. The main topics covered in this chapter include:

9.1 Introduction

In the previous chapter, you have studied about trigonometric ratios. In this chapter, you will be studying about some ways in which trigonometry is used in the life around you. **Trigonometry** is one of the most ancient subjects studied by scholars all over the world. As we have studied in Chapter 8, trigonometry was invented because its need arose in astronomy. In this chapter, we will see how trigonometry is used for finding the heights and distances of various objects, without actually measuring them.

9.2 Heights and Distances

The topic discusses the line of sight, angle of deviation, angle of elevation and angle of depression. All the processes are explained by solving some problems. The numerical problems are solved with the help of trigonometric ratios.

9.3 Summary

The summary describes all the points you have studied in the chapter. It will help you to understand the important concepts that need to be focused upon further from the chapter.

#### List of Exercises in Class 10 Maths Chapter 9 :

Exercise 9.1 Solutions โ 16 Questions ( 16 long answers)

In this chapter, โ**Some applications of trigonometry**โ, class 10 students will get to know how trigonometry is helpful in finding the height and distance of different objects without measuring. In earlier days, astronomers have used trigonometry for calculating the distance from the earth to the planets and stars. Trigonometry is mostly used in navigation and geography to locate the position in relation to the latitude and longitude.

Students will learn the applications of trigonometry with real-life examples in a better way. With the help of geometrical figures, the important terms and problems are explained and the summary is given at the end of the chapter. In NCERT Solutions for Class 10 Maths, you are provided with step by step procedure solutions of all the questions.

### Key Features of NCERT Solutions for Class 10 Maths Chapter 9 โ Some Applications of Trigonometry

- The solutions are designed by subject experts.
- Access for chapter-wise questions and answers.
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- The solutions are explained in the detailed procedure.
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Expert faculty team of members have formulated the solutions in a lucid manner to improve the problem solving abilities among the students. For a more clear idea about Some applications of trigonometry students can refer to the study materials available at BYJUโS.

## Frequently Asked Questions on NCERT Solutions for Class 10 Maths Chapter 9

### Where to download NCERT Solutions for Class 10 Maths Chapter 9?

NCERT Solutions for Class 10 Maths Chapter 9 can be downloaded at BYJUโS website. It can be avail in free PDF. To download go to BYJUโS NCERT Solutions website/select the Class 10/opt the subject as Maths/click on to the desired chapter that is Chapter 9 Some Applications of Trigonometry.

### What are the applications of learning NCERT Solutions for Class 10 Maths Chapter 9?

Applications of learning NCERT Solutions for Class 10 Maths Chapter 9 are finding the height and distance of different objects without measuring. By learning these concepts students will be able to answer all the questions based on trigonometry as well as it may help in writing class tests and board exams.

### Is NCERT Solutions for Class 10 Maths Chapter 9 enough for board exams?

Yes, you have to learn all the questions thoroughly and practice them for more time. Once you are done with solving all the questions then you can refer to other reference books and questions provided NCERT Exemplar textbooks.