Some Applications of Trigonometry

By
Dr. Upendra Kant Chaubey
-
0
300
NCERT Solutions for Class 10 Maths Chapter 9

NCERT Solutions for Class 10 Maths Chapter 9

Chapter 9 of NCERT Maths Class 10 delves into the practical applications of trigonometry in real-world scenarios. The key focus is on using trigonometric ratios to solve problems related to heights, distances, and angles.

Key Concepts and Applications:

  1. Line of Sight:
    • Angle of Elevation: The angle formed by the line of sight with the horizontal when looking upwards.
    • Angle of Depression: The angle formed by the line of sight with the horizontal when looking downwards.
  2. Height and Distance Problems:
    • Using trigonometric ratios to find the height of a building, tree, or other object, or the distance between two points.
  3. Practical Applications:
    • Surveying
    • Navigation
    • Engineering
    • Astronomy
    • Architecture

Key Trigonometric Ratios Used:

  • Sine (sin)
  • Cosine (cos)
  • Tangent (tan)

Problem-Solving Techniques:

  • Drawing diagrams to visualize the problem.
  • Identifying the appropriate trigonometric ratio based on the given information.
  • Setting up equations using trigonometric ratios.
  • Solving the equations to find the required values.

In essence, this chapter demonstrates the power of trigonometry in solving real-world problems involving height, distance, and angles. By understanding and applying trigonometric concepts, you can tackle various challenges in fields like engineering, surveying, and navigation.

NCERT Solutions for Class 10 Maths Chapter 9

Exercise 9.1

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°.

Ans : 

  • sin 30° = height of the pole / length of the rope

Substituting the given values:

  • sin 30° = height / 20 m

Since sin 30° = 1/2:

  • 1/2 = height / 20 m

Solving for height:

  • height = (1/2) * 20 m
  • height = 10 m

Therefore, the height of the pole is 10 meters.

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.

Ans : 

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?

Ans : 

Slide for younger children:

  • Height = 1.5 m
  • Angle of inclination = 30°

Using the sine function:

  • sin 30° = height / length
  • 1/2 = 1.5 / length
  • length = 1.5 * 2 = 3 m

Slide for elder children:

  • Height = 3 m
  • Angle of inclination = 60°

Using the sine function:

  • sin 60° = height / length
  • √3/2 = 3 / length
  • length = 3 * (2/√3)
  • length = 2√3 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 30°. Find the height of the tower.

Ans : 

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.

Ans : 

  • sin 60° = height / length of string

Substituting the given values:

  • √3/2 = 60 m / length of string

Solving for the length of the string:

  • length of string = 60 m * (2/√3)
  • length of string = (120√3) / 3
  • length of string = 40√3 m

Therefore, the length of the string is 40√3 meters.

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.

Ans : 

Given:

  • Height of the building  = 30 m
  • Height of the boy (AP) = 1.5 m
  • Angle of elevation from A to C = 30°
  • Angle of elevation from B to C = 60°

To find:

  • Distance AB (how far the boy walked)
Solution

Step 1: Find AC

  • From right triangle APC, tan 30° = PC/AC
  • AC = PC / tan 30° = (BC – BP) / tan 30°
  • AC = (30 – 1.5) / (1/√3) = 28.5 * √3 m

Step 2: Find BC

  • From right triangle BPC, tan 60° = PC/BC
  • BC = PC / tan 60° = (BC – BP) / √3
  • BC = (30 – 1.5) / √3 = 28.5 * √3 m

Step 3: Find AB

  • AB = AC – BC
  • AB = 28.5 * √3 – 28.5 * √3 / 3
  • AB = 28.5 * (√3 – 1/√3)
  • AB = 28.5 * (2√3 / 3)
  • AB = 19√3 m

Therefore, the distance the boy walked towards the building is 19√3 meters.

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.

Ans : 

  1. Find PA:
    • In right triangle PAB, tan 45° = AB/PA
    • 1 = 20/PA
    • PA = 20 m
  2. Find PC:
    • In right triangle PAC, tan 60° = AC/PA
    • √3 = (AB + BC)/20
    • √3 * 20 = AB + BC
    • BC = 20√3 – 20
  3. Find BC (height of the transmission tower):
    • BC = 20√3 – 20 m

Therefore, the height of the transmission tower is 20√3 – 20 meters.

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.

Ans : 

  1. Find PA:
    • In right triangle PAB, tan 45° = AB/PA
    • 1 = AB/PA
    • PA = AB
  2. Find PB:
    • In right triangle PBC, tan 60° = BC/PB
    • √3 = (AB + 1.6)/PB
    • PB = (AB + 1.6) / √3
  3. Equate PA and PB:
    • AB = (AB + 1.6) / √3
    • AB√3 = AB + 1.6
    • AB(√3 – 1) = 1.6
    • AB = 1.6 / (√3 – 1)
  4. Rationalize the denominator:
    • AB = 1.6 * (√3 + 1) / ((√3 – 1)(√3 + 1))
    • AB = 1.6 * (√3 + 1) / (3 – 1)
    • AB = 0.8 * (√3 + 1)

Height of the pedestal is 0.8(√3 + 1) meters.

9. The angle of elevation of the top of a building from the foot of a 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.

Ans : 

  1. Find PA:
    • In right triangle PAC, tan 60° = AC/PA
    • √3 = 50/PA
    • PA = 50/√3
  2. Find AB:
    • In right triangle PAB, tan 30° = AB/PA
    • 1/√3 = AB / (50/√3)
    • AB = 50/3

Therefore, the height of the building is 50/3 meters.

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 distance of the point from the poles.

Ans : 

  1. Find PA and PB:
    • In right triangle APA’, tan 60° = h/PA
    • √3 = h/PA
    • PA = h/√3
    • In right triangle BPB’, tan 30° = h/PB
    • 1/√3 = h/PB
    • PB = h√3
  2. Use the fact that AP + PB = 80 m:
    • h/√3 + h√3 = 80
    • h(1/√3 + √3) = 80
    • h(4/√3) = 80
    • h = 20√3

Therefore, the height of the poles is 20√3 meters.

  1. Find PA and PB:
    • PA = h/√3 = (20√3)/√3 = 20 m
    • PB = h√3 = 20√3 * √3 = 60 m

Therefore, the distance of the point P from pole A is 20 meters, and the distance of the point P from pole B is 60 meters.

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 the given figure). Find the height of the tower and the width of the CD and 20 m from pole AB.

NCERT Solutions For Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1 Q11

Ans : 

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.

Ans : 

  1. Find AD:
    • In right triangle ACD, tan 45° = AD/CD
    • 1 = AD/CD
    • AD = CD
  2. Find BD:
    • In right triangle ABD, tan 60° = BD/CD
    • √3 = BD/CD
    • BD = √3 * CD
  3. Find BC (height of the tower):
    • BC = BD + CD
    • BC = √3 * CD + CD
    • BC = CD(√3 + 1)
  4. Substitute CD = AD:
    • BC = AD(√3 + 1)
    • BC = 7(√3 + 1)

Therefore, the height of the tower is 7(√3 + 1) meters.

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.

Ans : 

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 60°. After sometime, the angle of elevation reduces to 30° (see figure). Find the distance travelled by the balloon during the interval.

NCERT Solutions For Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1 Q14

Ans : 

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.

Ans : 

NCERT Solutions for Class 10 Maths Chapter 9

FAQs

What is Chapter 9 of Class 10 Maths about?
Chapter 9 – Some Applications of Trigonometry explains how trigonometric ratios are used to find heights and distances in real-life situations.

Why is this chapter important for students?
It helps students understand practical uses of trigonometry and builds a foundation for higher-level maths and physics topics.

How do NCERT Solutions help in Chapter 9?
The NCERT Solutions provide step-by-step answers to all textbook problems, making it easier to understand and score well in exams.

What are the main topics covered in Chapter 9?
Key topics include line of sight, angle of elevation, angle of depression, and solving real-world height and distance problems.

Where can I get free NCERT Solutions for Class 10 Maths Chapter 9?
You can access free NCERT Solutions for Class 10 Maths Chapter 9 – Some Applications of Trigonometry – on reliable educational websites and apps.