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

Q: Q1. What are the most important topics in NCERT Class 10 Chapter 9 Some Applications of Trigonometry?

Answer: NCERT Solutions Class 10 Maths Chapter 9's main concepts Trigonometry applications include the angle of depression, angle of elevation, and line of sight. All questions based on these themes have detailed step-by-step responses. Students should make a point of drawing the corresponding figures after reading the question to better comprehend these topics. On the Toppr website, there is a comprehensive explanation for the answers.

Q: Q2. How many exercises are there in Chapter 9 Some Applications of Trigonometry of Class 10 Maths?

Answer: Exercise 9.1 at the end of Chapter 9 Some Applications of Trigonometry of Class 10 Maths is the only exercise in this chapter. This chapter's portion has 16 questions that are covered in the first two parts. There are a total of 16 questions in Class 10 Maths Chapter 9 Some Applications of Trigonometry, with 6 being quite easy, 5 being moderate, and 5 being long answer type problems.

Q: Q3. Explain a few real-life applications of trigonometry?

Answer: Today, trigonometry is utilized to refine difficult problems in many physical sciences and engineering topics. Trigonometry is employed in the development of computer music because sound travels in the form of waves, and this wave pattern employing sine and cosine functions aids in the development of computer music. Analytical trigonometry is essential in engineering fields such as mechanical engineering, electronics, and mechatronics. It has been used by astronomers to calculate distances from Earth to planets and stars, among other things. Trigonometry is also employed in geography and navigation. Trigonometry is used to make maps and calculate an island's position in reference to longitudes and latitudes. It is used in the marine and aviation industries. It is used in cartography (creation of maps). Trigonometry is also employed in satellite systems.

Question 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

^{o}

.

Easy

Solution

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Answer

Let AB be the vertical pole and CA be the rope. Then,

∠ A C B = 3 0^{o}

and

A C = 20 m

In right

△

ABC,

sin 3 0^{o} = \frac{A B}{A C}

\frac{1}{2} = \frac{A B}{2 0}

A B = 10 m

Therefore, the height of the pole is 10 m.

Answer

Let AB be the vertical pole and CA be the rope. Then,

∠ A C B = 3 0^{o}

and

A C = 20 m

In right

△

ABC,

sin 3 0^{o} = \frac{A B}{A C}

\frac{1}{2} = \frac{A B}{2 0}

A B = 10 m

Therefore, the height of the pole is 10 m.

Question 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 of

3 0^{o}

with the ground. 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.

Medium

Solution

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Answer

Let the Height of the Tree

= A B + A D

and given that

B D = 8 m

Now, when it breaks a part of it will remain perpendicular to the ground

(A B)

and remaining part

(A D)

will make an angle of

3 0^{o}

Now, in

△ A B D

cos 3 0^{o} = \frac{B D}{A D} \Rightarrow B D = \frac{3 A D}{2} \Rightarrow A D = \frac{2 \times 8}{3}

also, in the same Triangle

tan 3 0^{o} = \frac{A B}{B D} \Rightarrow A B = \frac{8}{3}

∴

Height of tree

= A B + A D = (\frac{1 6}{3} + \frac{8}{3}) m = \frac{2 4}{3} m = 83 m

Answer

Let the Height of the Tree

= A B + A D

and given that

B D = 8 m

Now, when it breaks a part of it will remain perpendicular to the ground

(A B)

and remaining part

(A D)

will make an angle of

3 0^{o}

Now, in

△ A B D

cos 3 0^{o} = \frac{B D}{A D} \Rightarrow B D = \frac{3 A D}{2} \Rightarrow A D = \frac{2 \times 8}{3}

also, in the same Triangle

tan 3 0^{o} = \frac{A B}{B D} \Rightarrow A B = \frac{8}{3}

∴

Height of tree

= A B + A D = (\frac{1 6}{3} + \frac{8}{3}) m = \frac{2 4}{3} m = 83 m

Question 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^{\circ}

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^{\circ}

to the ground. What should be the length of the slide in each case?

Medium

Solution

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Answer

Younger children

In

△ A B C

sin 3 0^{o} = \frac{B C}{A C}

\frac{1}{2} = \frac{1 . 5}{A C}

∴ A C = 3 m

Elder children

\frac{Y Z}{X Z} = \frac{3}{2} = \frac{3}{X Z}

X Z = \frac{6}{3} = 23 m

Answer

Younger children

In

△ A B C

sin 3 0^{o} = \frac{B C}{A C}

\frac{1}{2} = \frac{1 . 5}{A C}

∴ A C = 3 m

Elder children

\frac{Y Z}{X Z} = \frac{3}{2} = \frac{3}{X Z}

X Z = \frac{6}{3} = 23 m

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^{\circ}

. Find the height of the tower.

Medium

Solution

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Answer

In the above figure,

P R

is the tower, and

Q

is the point of observation at a distance of

30 m

from the base of the tower.

In

△ P Q R,

tan 3 0^{\circ} = \frac{P R}{P Q}

\Rightarrow P R = P Q tan 3 0^{\circ}

\Rightarrow P R = \frac{3 0}{3}

\Rightarrow P R = 103 m

∴

Height of the tower is

103 m

.

Answer

In the above figure,

P R

is the tower, and

Q

is the point of observation at a distance of

30 m

from the base of the tower.

In

△ P Q R,

tan 3 0^{\circ} = \frac{P R}{P Q}

\Rightarrow P R = P Q tan 3 0^{\circ}

\Rightarrow P R = \frac{3 0}{3}

\Rightarrow P R = 103 m

∴

Height of the tower is

103 m

.

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^{o}

. Find the length of the string assuming that there is no slack in the spring.

A

l = 602 m

B

l = 402 m

C

l = 202 m

D

None of these

Medium

Solution

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Answer

P

is the position of the kite. Its height from the point

Q

(on the ground)

= P Q = 60 m

Let

O P = l

be the length of the string

∠ P O Q = 60^{o}

(given)

s i n (Θ) = \frac{O p p o s i t e}{H y p o t e n u s e}

Now,

\frac{P Q}{O P} = sin 60^{o}

\Rightarrow

\frac{6 0}{l} = sin 60^{o} = \frac{3}{2}

\Rightarrow

\frac{6 0}{l} = \frac{3}{2}

\Rightarrow

l = \frac{3 \times 2 \times 2 0}{3} m

\Rightarrow

l = 403 m

The length of the string is

403 m

Answer

P

is the position of the kite. Its height from the point

Q

(on the ground)

= P Q = 60 m

Let

O P = l

be the length of the string

∠ P O Q = 60^{o}

(given)

s i n (Θ) = \frac{O p p o s i t e}{H y p o t e n u s e}

Now,

\frac{P Q}{O P} = sin 60^{o}

\Rightarrow

\frac{6 0}{l} = sin 60^{o} = \frac{3}{2}

\Rightarrow

\frac{6 0}{l} = \frac{3}{2}

\Rightarrow

l = \frac{3 \times 2 \times 2 0}{3} m

\Rightarrow

l = 403 m

The length of the string is

403 m

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 increase from

30^{\circ}

to

60^{\circ}

as he walks towards the building. Find the distance he walked towards the building.

Medium

Solution

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Answer

REF.Image

AB= 30-1.5=28.5 cm

In

Δ A B D : - t a n 3 0^{\circ} = \frac{A B}{B D}

\Rightarrow B D = (28.5) 3

In

Δ A B C : - t a n 6 0^{\circ} = \frac{A B}{B C}

\Rightarrow B C = \frac{( 2 8 . 5 )}{3}

CD= BD-BC

= 28.5 (3 - \frac{1}{3})

= 28.5 (\frac{2}{3}) = 32.91 m

Answer

REF.Image

AB= 30-1.5=28.5 cm

In

Δ A B D : - t a n 3 0^{\circ} = \frac{A B}{B D}

\Rightarrow B D = (28.5) 3

In

Δ A B C : - t a n 6 0^{\circ} = \frac{A B}{B C}

\Rightarrow B C = \frac{( 2 8 . 5 )}{3}

CD= BD-BC

= 28.5 (3 - \frac{1}{3})

= 28.5 (\frac{2}{3}) = 32.91 m

Question 7

From a point on the ground the angles of elevation of the bottom and top of a transmission tower fixed at the top of 20 m high building are

4 5^{o}

and

6 0^{o}

, respectively. Find the height of the transmission tower.

Medium

Solution

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Answer

Let DC be the tower and BC be the building. Then,

∠ C A B = 4 5^{o}, ∠ D A B = 6 0^{o}, B C = 20 m

Let height of the tower,

D C = h m

.

In right

△ A B C

,

tan 4 5^{o} = \frac{B C}{A B}

1 = \frac{2 0}{A B}

A B = 20 m

In right

△ A B D

,

tan 6 0^{o} = \frac{B D}{A B}

3 = \frac{h + 2 0}{2 0}

h = 20 (3 - 1) m

Answer

Let DC be the tower and BC be the building. Then,

∠ C A B = 4 5^{o}, ∠ D A B = 6 0^{o}, B C = 20 m

Let height of the tower,

D C = h m

.

In right

△ A B C

,

tan 4 5^{o} = \frac{B C}{A B}

1 = \frac{2 0}{A B}

A B = 20 m

In right

△ A B D

,

tan 6 0^{o} = \frac{B D}{A B}

3 = \frac{h + 2 0}{2 0}

h = 20 (3 - 1) m

Question 8

A statue 1.6 m tall stands on the top of pedestal. From a point on the 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.

Medium

Solution

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Answer

In

Δ B C D

tan 3 0^{o} = \frac{B C}{C D}

\frac{1}{3} = \frac{h}{C D}

C D = 3 h

m ........

(1)

Now,

In

Δ A C D

tan 6 0^{o} = \frac{A C}{C D}

= \frac{A B + B C}{C D}

From equation

(1)

substitution value of CD

tan 6 0^{o} = \frac{1 . 6 + h}{3 h}

\Rightarrow h 3 \times 3 = 1.6 + h

\Rightarrow 3 h - h = 1.6

2 h = 1.6

h = 0.8

m

Height of Pedestal.

Answer

In

Δ B C D

tan 3 0^{o} = \frac{B C}{C D}

\frac{1}{3} = \frac{h}{C D}

C D = 3 h

m ........

(1)

Now,

In

Δ A C D

tan 6 0^{o} = \frac{A C}{C D}

= \frac{A B + B C}{C D}

From equation

(1)

substitution value of CD

tan 6 0^{o} = \frac{1 . 6 + h}{3 h}

\Rightarrow h 3 \times 3 = 1.6 + h

\Rightarrow 3 h - h = 1.6

2 h = 1.6

h = 0.8

m

Height of Pedestal.

Question 9

The angle of elevation of the top of the building from the foot of the tower is

30

and the angle 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.

Medium

Solution

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Answer

Given height of tower

C D = 50 m

Let the height of the building,

A B = h

In right angled

△ B D C

,

\Rightarrow

tan 6 0^{o} = \frac{C D}{B D}

\Rightarrow

3 = \frac{5 0}{B D}

\Rightarrow

B D = \frac{5 0}{3} m

In right angled

△ A B D

,

\Rightarrow

tan 3 0^{o} = \frac{A B}{B D}

\Rightarrow

\frac{1}{3} = \frac{h}{\frac{5 0}{3}}

∴

\frac{1}{3} = \frac{3 h}{5 0}

∴

h = \frac{5 0}{3} = 16.66 m

∴

The height of the building is

16.66 m

.

Answer

Given height of tower

C D = 50 m

Let the height of the building,

A B = h

In right angled

△ B D C

,

\Rightarrow

tan 6 0^{o} = \frac{C D}{B D}

\Rightarrow

3 = \frac{5 0}{B D}

\Rightarrow

B D = \frac{5 0}{3} m

In right angled

△ A B D

,

\Rightarrow

tan 3 0^{o} = \frac{A B}{B D}

\Rightarrow

\frac{1}{3} = \frac{h}{\frac{5 0}{3}}

∴

\frac{1}{3} = \frac{3 h}{5 0}

∴

h = \frac{5 0}{3} = 16.66 m

∴

The height of the building is

16.66 m

.

Question 10

Two poles of equal heights are standing opposite to 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 top of the poles are

60^{o}

and

30^{o}

respectively. Find the height of the poles.

Medium

Solution

Verified by Toppr

Answer

Given that:

∠ A P B = 6 0^{\circ}, ∠ C P D = 3 0^{\circ}, A C = 80 m

To find:

The height of the pole

= A B = C D = ?

Solution:

Let

A B

and

C D

be the two poles of equal height and

P

be the point on the road between the poles.

In

△ A P B,

tan 6 0^{\circ} = \frac{A B}{A P}

or,

A P = A B \times \frac{1}{tan 6 0 ^{\circ}}

or,

A P = \frac{A B}{3}

- - - - - - - (i)

In

△ P C D,

tan 3 0^{\circ} = \frac{C D}{C P}

or,

C P = C D \times \frac{1}{tan 3 0 ^{\circ}}

or,

C P = 3 C D = 3 A B

∵ A B = C D

- - - - - - - (i i)

Adding eqn. (i) and eqn. (ii) we get,

A P + C P = \frac{A B}{3} + A B 3

or,

A C = A B (3 + \frac{1}{3})

or,

80 m = 4 \frac{A B}{3}

or,

A B = 203 m

Therefore, height of the pole

= 203 m = 34.64 m

Answer

Given that:

∠ A P B = 6 0^{\circ}, ∠ C P D = 3 0^{\circ}, A C = 80 m

To find:

The height of the pole

= A B = C D = ?

Solution:

Let

A B

and

C D

be the two poles of equal height and

P

be the point on the road between the poles.

In

△ A P B,

tan 6 0^{\circ} = \frac{A B}{A P}

or,

A P = A B \times \frac{1}{tan 6 0 ^{\circ}}

or,

A P = \frac{A B}{3}

- - - - - - - (i)

In

△ P C D,

tan 3 0^{\circ} = \frac{C D}{C P}

or,

C P = C D \times \frac{1}{tan 3 0 ^{\circ}}

or,

C P = 3 C D = 3 A B

∵ A B = C D

- - - - - - - (i i)

Adding eqn. (i) and eqn. (ii) we get,

A P + C P = \frac{A B}{3} + A B 3

or,

A C = A B (3 + \frac{1}{3})

or,

80 m = 4 \frac{A B}{3}

or,

A B = 203 m

Therefore, height of the pole

= 203 m = 34.64 m

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

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

NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry – Brief Overview

1. Applications of Trigonometry

2. Line of Sight and Angle of Elevation

Angle of Depression

3. Heights and Distances

Frequently Asked Questions on NCERT Class 10 Maths Chapter 9 : Some Applications of Trigonometry