## CBSE Class 10 Maths Notes Chapter 9 Some Applications of Trigonometry

Chapter 9 covers the application of trigonometry in solving real-world problems related to heights and distances. The chapter discusses various topics such as angles of elevation and depression, solving triangles, and the use of trigonometric ratios to find the height, distance, and angle of elevation or depression of an object.

Students will learn how to use the Pythagorean theorem, trigonometric ratios (sine, cosine, tangent), and inverse trigonometric functions (sine inverse, cosine inverse, tangent inverse) to solve problems related to heights and distances.

Some of the real-world problems discussed in this chapter include finding the height of a tower or building, the distance between two objects, and the angle of elevation or depression of an object.

**CBSE Class 10 Maths Important Questions Chapter 9 Some Applications of Trigonometry**

- Find the height of a tower if the angle of elevation of the top of the tower from a point on the ground is 60 degrees and the distance of the point from the tower is 20 meters.
- A ladder is leaning against a wall. The ladder is 10 meters long and is making an angle of 60 degrees with the ground. Find the distance of the base of the ladder from the wall.
- A kite is flying at a height of 45 meters from the ground. If the kite string is 60 meters long, find the angle that the string makes with the ground.
- The angle of elevation of the sun, when the length of the shadow of a tree is 3 times the height of the tree, is?
- A tower is 20 meters high. At a certain point, the angle of elevation of the top of the tower is 45 degrees. What is the distance of the point from the foot of the tower?
- A 15-meter-long ladder is placed against a wall. The foot of the ladder is 9 meters away from the wall. Find the height at which the ladder touches the wall.
- The angle of depression from the top of a building to a point on the ground 50 meters away from the building is 60 degrees. Find the height of the building.
- A 20-meter-high tower is situated on a hill. At a point on the ground, the angle of elevation of the top of the tower is 45 degrees. If the height of the hill is 50 meters, find the distance of the point from the foot of the hill.
- The angle of elevation of the top of a tower from a point on the ground, which is 30 meters away from the foot of the tower, is 45 degrees. Find the height of the tower.
- The angle of depression of a point on the ground from the top of a tower is 30 degrees. If the height of the tower is 20 meters, find the distance of the point from the foot of the tower.

Answers:

- 20√3 meters
- 5 meters
- 37 degrees
- 60 degrees
- 20 meters
- 12 meters
- 25 meters
- 10 meters
- 30 meters
- 20√3 meters

**CBSE Class 10 Maths Important Questions Answers Chapter 9 Some Applications of Trigonometry**

- In the given figure, ∠CAB = 90°, BC = 12 cm, AC = 20 cm, then find the length of AB.

Solution: We know that in a right-angled triangle, `sin θ = (Perpendicular/Hypotenuse)`

`cos θ = (Base/Hypotenuse)`

`tan θ = (Perpendicular/Base)`

Here, ∠CAB = 90° So, `sin CAB = sin 90° = 1`

Now, `sin CAB = (Perpendicular/Hypotenuse)`

So, `AB/AC = 1`

=> `AB = AC = 20 cm`

Therefore, the length of AB is 20 cm.

- Find the height of the pole, if a rope attached to its top makes an angle of 45° with the ground and the length of the rope is 12 m.

Solution: We know that in a right-angled triangle, `sin θ = (Perpendicular/Hypotenuse)`

`cos θ = (Base/Hypotenuse)`

`tan θ = (Perpendicular/Base)`

Here, the angle made by the rope with the ground is 45°.

So, `sin 45° = (Perpendicular/Hypotenuse)`

Now, we know that the length of the rope is 12 m. So, `Hypotenuse = 12 m`

Therefore, `Perpendicular = sin 45° × 12`

=> `Perpendicular = (1/√2) × 12`

=> `Perpendicular = 6√2 m`

Therefore, the height of the pole is 6√2 m.

- A ladder is placed against a wall such that it reaches the top of the wall, which is 5 m high. The ladder makes an angle of 60° with the ground. Find the length of the ladder.

Solution: We know that in a right-angled triangle, `sin θ = (Perpendicular/Hypotenuse)`

`cos θ = (Base/Hypotenuse)`

`tan θ = (Perpendicular/Base)`

Here, the angle made by the ladder with the ground is 60°.

So, `sin 60° = (Perpendicular/Hypotenuse)`

Now, we know that the height of the wall is 5 m. So, `Perpendicular = 5 m`

Therefore, `Hypotenuse = Perpendicular / sin 60°`

=> `Hypotenuse = 5 / sin 60°`

Now, `sin 60° = √3 / 2`

Therefore, `Hypotenuse = 5 / ( √3 / 2 )`

=> `Hypotenuse = (5 × 2) / √3`

=> `Hypotenuse = (10 / √3) m`

Therefore, the length of the ladder is 10/√3 m.

- A flagpole casts a shadow of 16 m when the angle of elevation of the sun is 45°. Find the height of the flagpole.

`sin θ = (Perpendicular/Hypotenuse)`

`cos θ = (Base/Hypotenuse)`

`tan θ = (Perpendicular/Base)`

Here, the angle of elevation of the sun is 45°. So, `sin 45° =

**CBSE Class 10 Maths Important Questions Answers MCQs Chapter 9 Some Applications of Trigonometry**

The angle of elevation of a tower from a point is 30°. If the distance of the point from the tower is 30 m, then the height of the tower is:

A) 15 m

B) 30√3 m

C) 30 m

D) 60 m

Answer: B) 30√3 m

In the given figure, PQ is parallel to BC. If AB = 5 cm, BC = 10 cm and AC = 7 cm, then the length of PQ is:

A) 5 cm

B) 7 cm

C) 10 cm

D) 14 cm

Answer: B) 7 cm

In the given figure, ∠BAC = 90° and AD is the altitude from A on BC. If AB = 5 cm, AC = 13 cm, and BD = 4 cm, then the value of CD is:

A) 8 cm

B) 9 cm

C) 10 cm

D) 11 cm

Answer: A) 8 cm

If sin A = 1/2 and cos B = 3/5, then the value of 2sin A cos B is:

A) 3/5

B) 2/5

C) 4/5

D) 1/5

Answer: B) 2/5

In the given figure, O is the center of the circle and ∠AOB = 90°. If AB = 24 cm and OD = 7 cm, then the length of BC is:

A) 17 cm

B) 21 cm

C) 23 cm

D) 25 cm

Answer: D) 25 cm

If cos θ = 3/5, then the value of sin θ is:

A) 4/5

B) 3/4

C) 5/4

D) 5/3

Answer: A) 4/5

The value of sin 45° is:

A) 1/2

B) 1/√2

C) √2/2

D) √3/2

Answer: C) √2/2

In a right-angled triangle ABC, right-angled at B, if AB = 12 cm and BC = 5 cm, then the value of sin A is:

A) 5/13

B) 12/13

C) 13/5

D) 13/12

Answer: A) 5/13

In the given figure, O is the center of the circle and AB is a diameter. If AC = 10 cm and BC = 24 cm, then the value of sin ∠CAB is:

A) 10/17

B) 15/17

C) 24/25

D) 7/25

Answer: B) 15/17

In a right-angled triangle ABC, right-angled at C, if tan A = 4/3, then the value of cos B is:

A) 3/5

B) 4/5

C) 5/3

D) 3/4

Answer: A) 3/5

## CBSE Class 10 Maths Notes Chapter 9 Some Applications of Trigonometry

**Line of Sight**

When an observer looks from a point E (eye) at object O then the straight line EO between eye E and object O is called the line of sight.

**Horizontal**

When an observer looks from a point E (eye) to another point Q which is horizontal to E, then the straight line, EQ between E and Q is called the horizontal line.

**Angle of Elevation**

When the eye is below the object, then the observer has to look up from point E to object O. The measure of this rotation (angle θ) from the horizontal line is called the angle of elevation.

**Angle of Depression**

When the eye is above the object, then the observer has to look down from point E to the object. The horizontal line is now parallel to the ground. The measure of this rotation (angle θ) from the horizontal line is called the angle of depression.

How to convert the above figure into the right triangle.**Case I:** Angle of Elevation is known

Draw OX perpendicular to EQ.

Now ∠OXE = 90°

ΔOXE is a rt. Δ, where

OE = hypotenuse

OX = opposite side (Perpendicular)

EX = adjacent side (Base)**Case II:** Angle of Depression is known

(i) Draw OQ’parallel to EQ

(ii) Draw perpendicular EX on OQ’.

(iii) Now ∠QEO = ∠EOX = Interior alternate angles

ΔEXO is an rt. Δ. where

EO = hypotenuse

OX = adjacent side (base)

EX = opposite side (Perpendicular)

- Choose a trigonometric ratio in such a way that it considers the known side and the side that you wish to calculate.
- The eye is always considered at ground level unless the problem specifically gives the height of the observer.

The object is always considered a point.**S**ome **P**eople **H**ave

Sin θ = PerpendicularHypotenuse**C**urly **B**lack **H**air

Cos θ = BaseHypotenuse**T**urning **P**ermanent **B**lack.

Tan θ = PerpendicularBase

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