NCERT Solutions for Class 10 Maths Chapter 9 Applications of Trigonometry

Our complete NCERT Solutions for Class 10 Maths Chapter 9 will help you learn how to measure the heights and distances of things in the real world.  Trigonometry isn't just about drawing triangles on paper; it's also about measuring the world around us. A lot of students have trouble picturing how the angles of elevation and depression operate in real life. This guide provides class 10 maths chapter 9 solutions to help you bridge the gap between theory and practical application.

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Applications of Trigonometry: Chapter Overview

Before you start the exercises, you need to know the main words used in class 10 maths chapter 9 solutions. These ideas are the basis for every problem you'll solve in this chapter.

• Line of Sight: This is the line that goes from the observer's eye to the spot on the object that the observer is looking at.

• Angle of Elevation: The angle of elevation is the angle between your line of sight and the horizontal level when you look up at anything.

• Angle of Depression: When you look down at something, the angle between the line of sight and the horizontal level is the angle of depression.

• Right-Angled Triangle: The foundation of all trigonometry, with one angle measuring exactly 90 degrees.

Important Trigonometric Ratios Table

To do well in class 10 maths NCERT solutions chapter 9, you need to memorise the values of certain angles. The table below shows the most common ratios:

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Applications of Trigonometry Class 10 NCERT Solutions

Here are the full answers to Exercise 9.1. These applications of trigonometry class 10 NCERT solutions focus on finding heights and distances using the tangent and sine ratios.

NCERT Solutions for Class 10 Maths Exercise 9.1

Solve the followings Questions.

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° (see figure ).

Answer:

Let AB be the vertical pole Ac be 20 m  long rope tied to point C. In  right ΔABC, sin 30° = AB/AC ⇒ 1/2 = AB/20 ⇒ AB = 20/2 ⇒ AB = 10 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 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 AC be the broken part of the tree. ∴ Total height of the tree = AB+AC In  right ΔABC, cos 30° = BC/AC ⇒ √3/2 = 8/AC ⇒ AC = 16/√3 Also, tan 30° = AB/BC ⇒ 1/√3 = AB/8 ⇒ AB = 8/√3 Total height of the tree = AB+AC = 16/√3 + 8/√3 = 24/√3 

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 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3 m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?

Answer:

There are two slides of height 1.5 m and 3 m. (Given) Let AB is 1.5 m and PQ be 3 m slides. ABC is the slide inclined at 30° with length AC and PQR is the slide inclined at 60° with length PR. A/q, In  right ΔABC, sin 30° = AB/AC ⇒ 1/2 = 1.5/AC ⇒ AC = 3m also,In  right ΔPQR, sin 60° = PQ/PR ⇒ √3/2 = 3/PR ⇒ PR = 2√3 m Hence, length of the slides are 3 m and 2√3 m respectively. 

Read More: NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

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 AB be the height of the tower and C is the point elevation which is 30 m away from the foot of the tower.In  right ΔABC, tan 30° = AB/BC ⇒ 1/√3 = AB/30 ⇒ AB = 10√3 Thus, the height of the tower is 10√3 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 60°. Find the length of the string, assuming that there is no slack in the string.

Answer:

Let BC be the height of the kite from the ground, AC be the inclined length of the string from the ground and A is the point where string of the kite is tied. A/q, In  right ΔABC, sin 60° = BC/AC ⇒ √3/2 = 60/AC ⇒ AC = 40√3 m Thus, the length of the string from the ground is 40√3 m. 

Read More: NCERT Solutions for Class 10 Maths Chapter 3

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 boy initially standing at point Y with inclination 30° and then he approaches the building to the point X with inclination 60°. ∴ XY is the distance he walked towards the building. also, XY = CD. Height of the building = AZ = 30 m AB = AZ - BZ = (30 - 1.5) = 28.5 m A/q, In  right ΔABD, tan 30° = AB/BD ⇒ 1/√3 = 28.5/BD ⇒ BD = 28.5√3 m also, In  right ΔABC, tan 60° = AB/BC ⇒ √3 = 28.5/BC ⇒ BC = 28.5/√3 = 28.5√3/3 m ∴ XY = CD = BD - BC = (28.5√3 - 28.5√3/3) = 28.5√3(1-1/3) = 28.5√3 × 2/3 = 57/√3 m. Thus, the distance boy walked towards the building is 57/√3 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 45° and 60° respectively. Find the height of the tower.

Answer:

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 - BC In  right ΔBCD, tan 45° = BC/CD ⇒ 1 = 20/CD ⇒ CD = 20 m also, In  right ΔACD, tan 60° = AC/CD ⇒ √3 = AC/20 ⇒ AC = 20√3 m Height of transmission tower = AB = AC - BC = (20√3 - 20) m = 20(√3 - 1) m. 

Read More: NCERT Solutions for Class 10 Maths Chapter 4

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 AB be the height of statue. D is the point on the ground from where the elevation is taken. Height of pedestal = BC = AC - AB In  right ΔBCD, tan 45° = BC/CD ⇒ 1 =  BC/CD ⇒ BC = CD. also, In  right ΔACD, tan 60° = AC/CD ⇒ √3 = AB+BC/CD ⇒ √3CD = 1.6 m + BC ⇒ √3BC = 1.6 m + BC ⇒ √3BC - BC = 1.6 m ⇒ BC(√3-1) = 1.6 m ⇒ BC = 1.6/(√3-1) m ⇒ BC = 0.8(√3+1) m Thus, the height of the pedestal is 0.8(√3+1) m. 

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 AB be the building and CD be the tower In ΔCDB, CB/BD = tan 60º 50/BD = √3 BD = √3/50 In ΔABD, (AB)/(BD) = tan 30º AB = 50/√3 x 1/√3 = 50/3 = 16 2/3 Therefore, the height of the building is 16 2/3 m. 

Read More: NCERT Solutions for Class 10 Maths Chapter 10

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:

In right triangle PRQ,Let AB and CD be the poles and O is the point from where the elevation angles are measured. In ΔABO (AB)/(BO) = tan 60º AB/BO = √3 BO = AB/√3 In ΔCDO, (CD)/(DO) = tan 30º CD/80 - BO = 1/√3 CD x √3 = 80 - BO CD x √3 = 80 - AB/√3 CD x √3 + AB/√3 = 80 Since the poles are of equal heights, CD = AB CD[√3 + 1/√3] = 80 CD[3+1/√3] = 80 CD = 20√3 BO = AB/√3 = CD/√3 = [20√3/√3]m = 20m DO = BD − BO = (80 − 20) m = 60 m Hence the heights of the poles are 20√3 m each and the distances of the point from poles 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 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° (see figure). Find the height of the tower and the width of the canal.

Answer:

In ΔABC, AB/BC = tan 60º AB/BC = √3 BC = AB/ √3 In ΔABD, AB/BD = tan 30º AB/BC+CD = 1/√3 [AB/(AB/√3) + 20] = 1/√3 [AB x √3/AB + 20 x √3] = 1/√3 3AB = AB + 20√3 = 2AB = 20√3 AB = 10√3m BC = AB/√3 = {10√3/√3}m = 10m Hence height of the tower is 10√3 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 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

Answer:

Let AB be a building and CD be a cable tower. In ΔABD, AB/BD = tan 45º 7/BD = 1 BD = 7 m In ΔACE, AE = BD = 7 m CE/AE = tan 60º CE/7 = √3 CE = 7 x √3 CD = CE + ED = [7 x √3 + 7]m = 7[√3 + 1]m Hence height of the tower is 7[√3 + 1]m. 

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 AB be the lighthouse and the two ships be at point C and D respectively. In ΔABC, AB/BC = tan 45º 75/BC = 1 BC = 75 m In ΔABD, AB/BD = tan 30º 75/BC+CD = 1/√3 75/75+CD = 1/√3 75 x √3 = 75 + CD 75[√3 - 1]m = CD Hence the distance between the two ships is 75[√3 - 1]m 

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 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.

Answer:

In right triangle ABC,Let the initial position A of balloon change to B after some time and CD be the girl. In ΔACE, AE/CE = tan 60º (AF - EF)/(CE) = tan 60º 88.2 - 1.2/CE = √3 87/CE = √3 CE = 87/√3 = 87x√3m In ΔBCG, (BG)/(CG) = tan 30º 88.2 - 1.2/CG = 1/√3 87/CG = 1/√3 CG = 87x√3 m Distance travelled by balloon = EG = CG − CE = (87x√3 - 29x√3 )m = 58√3m Hence the distance travelled by the balloon during the interval is 58√3 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 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 AB be the tower. Initial position of the car is C, which changes to D after six seconds. In ΔADB, AB/DB = tan 60º AB/DB = √3 DB = AB/√3 In ΔABC, AB/BC = tan 30º AB/BD + DC = 1/√3 AB √3 = BD + DC AB √3 = AB/√3 + DC DC = AB √3 - AB/√3 = AB(√3 - 1/√3) = 2AB/√3 Time taken by the car to travel a distance DC (i.e.2AB/√3) = 6 seconds. Time taken by the car to travel a distance DB (i.e. AB/√3) = [6/2AB/√3] x [AB/√3] = 6/2 = 3 seconds. Hence, the further time taken by the car to reach the foot of the tower is 3 seconds. 

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.

Answer:

Let AQ be the tower and R, S are the points 4m, 9m away from the base of the tower respectively. The angles are complementary. Therefore, if one angle is θ, the other will be 90 − θ. In ΔAQR, AQ/QR = tanΘ AQ/4 = tanΘ    ... 1 In ΔAQS, AQ/SQ = tan(90 - Θ) AQ/9 = cot Θ ...2 On multiplying equations (1) & (2) (AQ/4)(AQ/9) = (tanΘ).(cot Θ) AQ²/36 = 1 AQ² = 36 AQ = √36 AQ = ±6 However, height cannot be negative. Therefore, the height of the tower is 6 m

Read More: NCERT Solutions for Class 10 Maths Chapter 14

Extra Practice Questions for Class 10 Applications of Trigonometry

As we progress through, the problems involve more variables and complex visualisations.

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.

Solution:

1. Let AB be the tower and C be the point on the ground.

2. BC = 30 m and 1\angle ACB = 30°.

3. In right triangle ABC, 1\tan 30° = AB / BC.

4. 1/\sqrt{3} = AB / 30.

5. AB = 30/\sqrt{3} = 10\sqrt{3} m.
   The height of the tower is 10\sqrt{3} metres.

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 there is no slack in the string.

Solution:

1. Let K be the kite and P be the point on the ground.

2. Height HK = 60 m. Angle = 60°.

3. \sin 60° = \text{Height} / \text{String Length}$.

4. 1\sqrt{3}/2 = 60 / L.

5. L = 120/\sqrt{3} = 40\sqrt{3}$m.
   The length of the string is 40\sqrt{3} metres.

When dealing with application to trigonometry class 10 solutions, you often encounter two different angles of observation. These are usually the most weightage-carrying questions in exams.

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.

Solution:

1. Height of building = 30 m. Height of boy = 1.5 m.

2. Adjusted height (AB) = 30 - 1.5 = 28.5 m.

3. In the first position, 1\tan 30° = 28.5 / x \implies x = 28.5\sqrt{3}.

4. In the second position, 1/tan 60° = 28.5 / y \implies y = 28.5/\sqrt{3}.

5. Distance walked = x - y = 28.5\sqrt{3} - 28.5/\sqrt{3}.

6. Distance = 28.5(3-1)/\sqrt{3} = 57/\sqrt{3} = 19\sqrt{3} m.
   The boy walked 19\sqrt{3} metres towards the building.

By practising these class 10 maths chapter 9 solutions, you develop a strong spatial intuition that is useful not just in maths, but also in physics.

Summary of Formulae

For students' convenience, here's a short summary: 

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NCERT Solutions for Class 10 Maths Chapter 9 FAQs

What is the main use of class 10 maths chapter 9 solutions?

These answers assist students learn how to use trigonometric ratios to figure out how far apart things are and how tall buildings and mountains are without actually measuring them.

Which ratio is most common in class 10 maths NCERT solutions chapter 9?

The tangent ($\tan$) ratio is the most common one because most issues ask for either the height (perpendicular) or the distance (base) of an item.

Is drawing a diagram mandatory in trigonometry?

Yes, in CBSE exams, drawing a proper and labelled diagram for class 10 mathematics chapter 9 solutions is worth a lot of points and helps you avoid making mistakes when you do the arithmetic.

Do you need to know the values of $\sqrt{2}$ and $\sqrt{3}$ to use trigonometry?

While many answers are accepted in root form, knowing that $\sqrt{3} \approx 1.732$ and $\sqrt{2} \approx 1.414$ is helpful for objective questions or specific decimal requirements.

How can I master cbse class 10 maths some applications of trigonometry solutions?

The best way is to solve every NCERT exercise problem and focus on identifying whether the angle given is an angle of elevation or depression.