# Some Applications Of Trigonometry Chapter 9 Class 10 Maths NCERT PDF

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### Chapter 9: Some Applications Of Trigonometry

9.1 Introduction

In the previous chapter, you have studied trigonometric ratios. In this chapter, you will be studying 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 said in Chapter 8, trigonometry was invented because its need arose in astronomy.

Since then astronomers have used it, for instance, to calculate distances from the Earth to the planets and stars.

Trigonometry is also used in geography and in navigation. The knowledge of trigonometry is used to construct maps, and determine the position of an island in relation to the longitudes and latitudes.

#### 9.3 Summary

In this chapter, you have studied the following points :

1. (i) The line of sight is the line drawn from the eye of an observer to the point in the object viewed by the observer.
(ii) The angle of elevation of an object viewed, is the angle formed by the line of sight with the horizontal when it is above the horizontal level, i.e., the case when we raise our head to look at the object.
(iii) The angle of depression of an object viewed, is the angle formed by the line of sight with the horizontal when it is below the horizontal level, i.e., the case when we lower our head to look at the object.
2. The height or length of an object or the distance between two distant objects can be determined with the help of trigonometric ratios.

### NCERT Solutions Class 10 Maths Chapter 9 Some Applications Of Trigonometry

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 fig. 9.11)

Solution:

The length of the rope is 20 m and the angle made by the rope with the ground level is 30°.

Given: AC = 20 m and angle C = 30°

To Find: Height of the pole

Let AB be the vertical pole

In right ΔABC, using the sine formula

sin 30° = AB/AC

Using the value of sin 30 degrees is ½, 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 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.

Solution:

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

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 ΔABC,

Using Cosine and tangent angles,

cos 30° = BC/AC

We know that, cos 30° = √3/2

√3/2 = 8/AC

AC = 16/√3 …(1)

Also,

tan 30° = AB/BC

1/√3 = AB/8

AB = 8/√3 ….(2)

Therefore, total height of the tree = AB + AC = 16/√3 + 8/√3 = 24/√3 = 8√3 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 30° 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 60° to the ground. What should be the length of the slide in each case?

Solution:

As per the contractor’s plan,

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

60° with length PR.

To Find: AC and PR

In right ΔABC,

sin 30° = AB/AC

1/2 = 1.5/AC

AC = 3

Also,

In right ΔPQR,

sin 60° = PQ/PR

⇒ √3/2 = 3/PR

⇒ PR = 2√3

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

Length of the slide for elders children = 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.

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 30° = AB/BC

1/√3 = AB/30

⇒ AB = 10√3

Thus, the height of the tower is 10√3 m.