# Class 10 NCERT Solutions- Chapter 9 Some Application of Trigonometry – Exercise 9.1

Last Updated : 03 Apr, 2024

### 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 30Â° (see Fig.).

Solution:

In rt âˆ†ABC,

AB = pole = ?

AC = rope = 20m

sinÎ¸ = [Tex]\frac{P}{H}[/Tex]

sin30Â° = [Tex]\frac{AB}{AC}[/Tex]

AB = 1/2 * 20

AB = 10m

Height of pole = 10m

### 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 30Â° 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:

In rt âˆ†ABC,

BC = 8m

[Tex]\frac{P}{B} = \frac{AB}{BC}    [/Tex] = tan30Â°

[Tex]\frac{AB}{BC}    [/Tex] = 1/âˆš3

AB = 8/âˆš3          -(1)

Now,

[Tex]\frac{BC}{ AC}    [/Tex] = cos30Â°

8/AC = âˆš3/2

âˆš3AC = 16

AC = 16/âˆš3           -(2)

From (1) and (2)

Height of tree = AB + AC

= 8/âˆš3 * 16âˆš3

= 8âˆš3 m

8 * 1.73 = 13.84m

The height of the tree is 13.84

### 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 30Â° 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 60Â° to the ground. What should be the length of the slide in each case?

Solution:

In rt âˆ†ABC,

AB = 1.5m

AC = side = ?

[Tex]\frac{P}{H} = \frac{AB}{AC}    [/Tex] = sin30Â°

1.5/AC = 1/2

AC = 1/5 * 2

AC = 3m

In rt âˆ†PQR,

PQ = 3m

PR = side = ?

[Tex]\frac{P}{H} = \frac{PQ}{PR}    [/Tex] = sin60Â°

3/PR = âˆš3/2

âˆš3 PR = 6

PR = 6/âˆš3

6/âˆš3 * âˆš3/âˆš3

= 2âˆš3

= 2 * 1.73

= 3.46m

### 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 30Â°. Find the height of the tower.

Solution:

In rt âˆ†ABC,

AB = tower = ?

BC = 30m

[Tex]\frac{P}{B} = \frac{AB}{BC}    [/Tex] = tan30Â°

AB/30 = 1/âˆš3

AB = 30/âˆš3

AB = 30/âˆš3 * âˆš3/âˆš3

= (30âˆš3)/3 = 10âˆš3

= 10 * 1.73

= 17.3m

The height of tower 17.3m

### 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 60Â°. Find the length of the string, assuming that there is no slack in the string.

Solution:

In rt âˆ†ABC,

AB = 6Om

AC = string = ?

[Tex]\frac{P}{H} = \frac{AB}{AC}     [/Tex]= sin60Â°

60/AC = âˆš3/3

âˆš3 AC = 60 * 2

AC = 120/120/(âˆš3) * âˆš3/âˆš3

120/âˆš3 * âˆš3/âˆš3

40 = âˆš3

40 * 1.73 = 69.20m

Length of the string is 69.20m

### 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 30Â° to 60Â° as he walks towards the building. Find the distance he walked towards the building.

Solution:

In fig AB = AE – 1.5

= 30 – 1.5

= 28.5

In rt âˆ†ABD,

[Tex]\frac{P}{B} = \frac{AB}{BD}    [/Tex] = tan30Â°

= 28.5/BD = 1/âˆš3

BD = 28.5âˆš3           -(1)

In rt âˆ†ABC,

[Tex]\frac{P}{B} = \frac{AB}{BC}    [/Tex] = tan60Â°

28.5/BC*âˆš3

âˆš3 BC = 28.5

BC = 28.5/âˆš3          -(2)

CD = BD âˆ’ BC

= 28.5âˆš3 – 28.5/âˆš3

= 28.5(2/âˆš3)

57/âˆš3 * âˆš3/âˆš3 = (57âˆš3)/3 = 19âˆš3

19 * 1.73 = 32.87m

The boy walked 32.87m towards the building.

### Question 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 45Â° and 60Â° respectively. Find the height of the tower.

Solution:

In fig:

AB = tower = ?

BC = building = 20m

In rt âˆ†BCD

[Tex]\frac{P}{B} = \frac{BC}{CD}    [/Tex] = tan45Â°

20/CD = 1/1

CD = 20

In rt. âˆ†ACD,

[Tex]\frac{P}{B} = \frac{AC}{CD}     [/Tex]= tan60Â°

AC/20 = âˆš3/1

AC = 20âˆš3           -(1)

AB = AC-BC

20âˆš3 – 20

20(âˆš3 – 1)

20(1.732 – 1)

20(0.732)

14.64m

The height of the tower is 14.6m

### Question 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 60Â° and from the same point, the angle of elevation of the top of the pedestal is 45Â°. Find the height of the pedestal.

Solution:

In fig: AB = statue = 1.6m

BC = pedestal = ?

In rt âˆ†ACD

[Tex]\frac{P}{B} = \frac{AC}{CD}    [/Tex] = tan60Â°

1.6 + BC/CD = âˆš3

âˆš3 CD = 1.6 + BC

CD = 1.6+BC/âˆš3           -(1)

In rt âˆ†BCD,

[Tex]\frac{BC}{CD}    [/Tex] = tan45Â°

[Tex]\frac{BC}{CD}    [/Tex] = 1/1

CD = BC

From (1)

1.6 + BC/âˆš3 = BC/1

âˆš3 BC = 1.6 + BC

1.732 BC – 1 BC = 1.6

0.732 * BC = 1.6

BC = 1.6/0.732

BC = 16/10 * 100/732 = 1600/732

BC = 2.18m

Height of pedestal is 2.18m

### Question 9. The angle of elevation of the top of a building from the foot of the tower is 30Â° and the angle of elevation of the top of the tower from the foot of the building is 60Â°. If the tower is 50 m high, find the height of the building.

Solution:

In fig:

AB = tower = 50m

DC = building = ?

In rt.âˆ†ABC,

[Tex]\frac{P}{B} = \frac{AB}{BC}    [/Tex] = tan60Â°

âˆš3 BC = 50

BC = 50/âˆš3

In rt. âˆ†DCB

[Tex]\frac{P}{B} = \frac{DC}{BC}    [/Tex] = tan30Â°

[Tex]\frac{P}{B} = \frac{DC}{\frac{50}{âˆš3}}    [/Tex] = 1/âˆš3

DC = 50/âˆš3

DC = 50/âˆš3 * 1/âˆš3

DC = 50/3

DC = [Tex] 16\tfrac{2}{3}[/Tex]

The height of the building is [Tex] 16\tfrac{2}{3}     [/Tex]m

### Question 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 60Â° and 30Â°, respectively. Find the height of the poles and the distances of the point from the poles.

Solution:

AB and CD on equal poles.

Let their height = h

Let DP = x

Then PB = BD – x

In rt. âˆ†CDP,

[Tex]\frac{P}{B} = \frac{CD}{DP}    [/Tex] = tan60Â°

h/x = âˆš3/1

h = âˆš3 x           -(1)

In rt. âˆ†ABP

[Tex]\frac{P}{B} = \frac{AB}{BP}    [/Tex] = tan30Â°

h/(80 – x) = 1/âˆš3

h = (80 – x)/âˆš3          -(2)

From (1) and (2)

(âˆš3 x)/1 = 80 – x/âˆš3

3x = 80 – x

3x + x = 80

4x = 80

X = 80/4

X = 20

Putting values of X in equation 1

h = âˆš3 x

h = âˆš3(20)

h = 1.732(20)

h = 34.640

Height of each pole = 34.64m

The point is 20m away from first pole and 60m away from second pole.

### Question 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 60Â°. 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 30Â° (see Fig.). Find the height of the tower and the width of the canal.

Solution:

In fig: AB = tower = ?

CB = canal = ?

In rt. âˆ†ABC,

tan60Â° = [Tex]\frac{AB}{BC} [/Tex]

h/x = âˆš3

h = âˆš3 x          -(1)

In rt. âˆ†ABD

[Tex]\frac{AB}{BD}   [/Tex] = tan 30Â°

[Tex]\frac{h}{x + 20}   [/Tex] = 1/âˆš3

h = (x + 20)/âˆš3          -(2)

From (1) and (2)

âˆš3/1 = (x + 20)/âˆš3

3x = x + 20

3x – x = 20

2x = 20

X = 20/2

X = 10

Width of the canal is 10m

Putting value of x in equation 1

h = âˆš3 x

= 1.732(10)

= 17.32

Height of the tower 17.32m.

### Question 12. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60Â° and the angle of depression of its foot is 45Â°. Determine the height of the tower.

Solution:

In fig: ED = building = 7m

AC = cable tower = ?

In rt âˆ†EDC,

[Tex]\frac{ED}{DC}   [/Tex] = tan45Â°

7/x = 1/1

DC = 7

Now, EB = DC = 7m

In rt. âˆ†ABE,

[Tex]\frac{AB}{BC}   [/Tex] = tan60Â°

AB/7 = âˆš3/1

Height of tower = AC = AB + BC

7âˆš3 + 7

= 7(âˆš3 + 1)

= 7(1.732 + 1)

= 7(2.732)

Height of cable tower = 19.125m

### Question 13. As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30Â° and 45Â°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Solution:

In fig:

AB = lighthouse = 75m

D and C are two ships

DC = ?

In rt. âˆ†ABD,

[Tex]\frac{AB}{BD}   [/Tex] = tan30Â°

75/BD = 1/âˆš3

BD = 75âˆš3

In rt. âˆ†ABC

[Tex]\frac{AB}{BC}   [/Tex] = tan45Â°

75/BC = 1/1

BC = 75

DC = BD – BC

= 75âˆš3 – 75

75(âˆš3 – 1)

75(1.372 – 1)

34.900

Hence, distance between two sheep is 34.900

### Question 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 60Â°. After some time, the angle of elevation reduces to 30Â° (see Fig.). Find the distance traveled by the balloon during the interval.

Solution:

In fig: AB = AC – BC

= 88.2 – 1.2

= 81m

In rt. âˆ†ABE

[Tex]\frac{AB}{EB}   [/Tex] = 87/EB = tan30Â°

87/EB = 1/âˆš3

EB = 87âˆš3

In rt. âˆ†FDE

[Tex]\frac{FD}{ED}   [/Tex] = tan60Â°

âˆš3 ED = 87

ED = 87/âˆš3

DB = DB – ED

87âˆš3 – 87/âˆš3

87(âˆš3 – 1/âˆš3)

= 87(3 – 1/âˆš3)

= 87(2/âˆš3) = 174/âˆš3 * âˆš3/âˆš3

= 174 * âˆš3/3 = 58âˆš3

58 * 1.732 = 100.456m

Distance traveled by balloon is 100.456m

### Question 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 30Â°, 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 60Â°. Find the time taken by the car to reach the foot of the tower from this point.

Solution:

In fig: AB is tower

In rt. âˆ†ABD

[Tex]\frac{AB}{DB}   [/Tex] = tan30Â°

[Tex]\frac{AB}{DB}   [/Tex] = 1/âˆš3

DB = âˆš3 AB          -(1)

In rt. âˆ†ABC

[Tex]\frac{AB}{BC}   [/Tex] = tan60Â°

BC = AB/âˆš3          -(2)

DC = DB – BC

= âˆš3 AB – AB/âˆš3

AB(3 – 1/âˆš3)

CD = 2AB/âˆš3

[Tex]Speed = \frac{Distance}{Time}[/Tex]

S1 = S2

\frac{D1}{T1} = \frac{D2}{T2}

[Tex]\frac{DC}{6} = \frac{CB}{t}[/Tex]

2/âˆš3AB/6 = AB/âˆš3/t

2t = 6

t = 6/2

t = 3sec

### Question 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.

Solution:

In fig: AB is tower

To prove: AB = 6m

Given: BC = 4m          DB = 9m

In âˆ†ABC

[Tex]\frac{AB}{BC}   [/Tex] = tanÎ¸

AB/4 = tanÎ¸         -(1)

In âˆ†ABD

[Tex]\frac{AB}{BD}   [/Tex] = tan (90Â°-Î¸)

AB/9 = 1/ tanÎ¸

9/AB = tanÎ¸         -(2)

From (1) and (2)

AB/4 = 9/AB

AB2 = 36

AB = âˆš36

AB = âˆš(6 * 6)

AB = 6m

Height of the tower is 6m.

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