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

Last Updated : 03 Apr, 2024
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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° (see Fig.).  

Solution:

In rt ∆ABC,

AB = pole = ?

AC = rope = 20m

sinθ = [Tex]\frac{P}{H}[/Tex]

sin30° = [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 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:

In rt ∆ABC,

BC = 8m

[Tex]\frac{P}{B} = \frac{AB}{BC}    [/Tex] = tan30°

[Tex]\frac{AB}{BC}    [/Tex] = 1/√3

AB = 8/√3          -(1)

Now,

[Tex]\frac{BC}{ AC}    [/Tex] = cos30°

8/AC = √3/2

√3AC = 16  

AC = 16/√3           -(2)

From (1) and (2)

Height of tree = AB + AC

= 8/√3 * 16√3

= 8√3 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 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?

Solution:

In rt ∆ABC,

AB = 1.5m

AC = side = ?

[Tex]\frac{P}{H} = \frac{AB}{AC}    [/Tex] = sin30°

1.5/AC = 1/2

AC = 1/5 * 2

AC = 3m

In rt ∆PQR,

PQ = 3m

PR = side = ?

[Tex]\frac{P}{H} = \frac{PQ}{PR}    [/Tex] = sin60°

3/PR = √3/2

√3 PR = 6  

PR = 6/√3

6/√3 * √3/√3

= 2√3

= 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 30°. Find the height of the tower.

Solution:

In rt ∆ABC,

AB = tower = ?

BC = 30m

[Tex]\frac{P}{B} = \frac{AB}{BC}    [/Tex] = tan30°

AB/30 = 1/√3

AB = 30/√3

AB = 30/√3 * √3/√3  

= (30√3)/3 = 10√3  

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

Solution:

In rt ∆ABC,

AB = 6Om

AC = string = ?

[Tex]\frac{P}{H} = \frac{AB}{AC}     [/Tex]= sin60°

60/AC = √3/3

√3 AC = 60 * 2  

AC = 120/120/(√3) * √3/√3

120/√3 * √3/√3  

40 = √3  

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 30° to 60° 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 ∆ABD,

[Tex]\frac{P}{B} = \frac{AB}{BD}    [/Tex] = tan30°

= 28.5/BD = 1/√3

BD = 28.5√3           -(1)

In rt ∆ABC,

[Tex]\frac{P}{B} = \frac{AB}{BC}    [/Tex] = tan60°

28.5/BC*√3  

√3 BC = 28.5

BC = 28.5/√3          -(2)   

CD = BD − BC

= 28.5√3 – 28.5/√3          

= 28.5(2/√3)        

57/√3 * √3/√3 = (57√3)/3 = 19√3  

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 45° and 60° respectively. Find the height of the tower.

Solution:

In fig:

AB = tower = ?

BC = building = 20m

In rt ∆BCD

[Tex]\frac{P}{B} = \frac{BC}{CD}    [/Tex] = tan45°

20/CD = 1/1

CD = 20

In rt. ∆ACD,

[Tex]\frac{P}{B} = \frac{AC}{CD}     [/Tex]= tan60°

AC/20 = √3/1

AC = 20√3           -(1)   

AB = AC-BC

20√3 – 20

20(√3 – 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 60° and from the same point, the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.

Solution:

In fig: AB = statue = 1.6m

BC = pedestal = ?

In rt ∆ACD

[Tex]\frac{P}{B} = \frac{AC}{CD}    [/Tex] = tan60°

1.6 + BC/CD = √3

√3 CD = 1.6 + BC

CD = 1.6+BC/√3           -(1) 

In rt ∆BCD,  

[Tex]\frac{BC}{CD}    [/Tex] = tan45°  

[Tex]\frac{BC}{CD}    [/Tex] = 1/1  

CD = BC

From (1) 

1.6 + BC/√3 = BC/1

√3 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 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.

Solution:

In fig:

AB = tower = 50m

DC = building = ?

In rt.∆ABC,

[Tex]\frac{P}{B} = \frac{AB}{BC}    [/Tex] = tan60°

√3 BC = 50  

BC = 50/√3

In rt. ∆DCB

[Tex]\frac{P}{B} = \frac{DC}{BC}    [/Tex] = tan30° 

[Tex]\frac{P}{B} = \frac{DC}{\frac{50}{√3}}    [/Tex] = 1/√3  

DC = 50/√3

DC = 50/√3 * 1/√3  

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 60° and 30°, 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. ∆CDP,

[Tex]\frac{P}{B} = \frac{CD}{DP}    [/Tex] = tan60°

h/x = √3/1

h = √3 x           -(1)

In rt. ∆ABP

[Tex]\frac{P}{B} = \frac{AB}{BP}    [/Tex] = tan30°

h/(80 – x) = 1/√3

h = (80 – x)/√3          -(2)

From (1) and (2)

(√3 x)/1 = 80 – x/√3  

3x = 80 – x

3x + x = 80

4x = 80

X = 80/4

X = 20

Putting values of X in equation 1

h = √3 x

h = √3(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.

Chapter 9 Some Application of Trigonometry – Exercise 9.1 | Set 1

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 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 Fig.). Find the height of the tower and the width of the canal.

Solution:

In fig: AB = tower = ?

CB = canal = ?

In rt. ∆ABC,

tan60° = [Tex]\frac{AB}{BC} [/Tex]

h/x = √3

h = √3 x          -(1)

In rt. ∆ABD

[Tex]\frac{AB}{BD}   [/Tex] = tan 30°

[Tex]\frac{h}{x + 20}   [/Tex] = 1/√3

h = (x + 20)/√3          -(2)

From (1) and (2)

√3/1 = (x + 20)/√3  

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 = √3 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 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

Solution:

In fig: ED = building = 7m

AC = cable tower = ?

In rt ∆EDC,

[Tex]\frac{ED}{DC}   [/Tex] = tan45°

7/x = 1/1

DC = 7

Now, EB = DC = 7m

In rt. ∆ABE,

[Tex]\frac{AB}{BC}   [/Tex] = tan60°

AB/7 = √3/1

Height of tower = AC = AB + BC

7√3 + 7

= 7(√3 + 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 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.

Solution:

In fig:

AB = lighthouse = 75m

D and C are two ships

DC = ?

In rt. ∆ABD,

[Tex]\frac{AB}{BD}   [/Tex] = tan30°

75/BD = 1/√3

BD = 75√3

In rt. ∆ABC

[Tex]\frac{AB}{BC}   [/Tex] = tan45°

75/BC = 1/1

BC = 75

DC = BD – BC

= 75√3 – 75

75(√3 – 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 60°. After some time, the angle of elevation reduces to 30° (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. ∆ABE

[Tex]\frac{AB}{EB}   [/Tex] = 87/EB = tan30°

87/EB = 1/√3

EB = 87√3

In rt. ∆FDE

[Tex]\frac{FD}{ED}   [/Tex] = tan60°

√3 ED = 87  

ED = 87/√3

DB = DB – ED

87√3 – 87/√3

87(√3 – 1/√3)

= 87(3 – 1/√3)

= 87(2/√3) = 174/√3 * √3/√3

= 174 * √3/3 = 58√3

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 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.

Solution:

In fig: AB is tower

In rt. ∆ABD  

[Tex]\frac{AB}{DB}   [/Tex] = tan30°

[Tex]\frac{AB}{DB}   [/Tex] = 1/√3

DB = √3 AB          -(1)

In rt. ∆ABC  

[Tex]\frac{AB}{BC}   [/Tex] = tan60°

BC = AB/√3          -(2)

DC = DB – BC

= √3 AB – AB/√3  

AB(3 – 1/√3)

CD = 2AB/√3

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

S1 = S2

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

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

2/√3AB/6 = AB/√3/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 ∆ABC

[Tex]\frac{AB}{BC}   [/Tex] = tanθ

AB/4 = tanθ         -(1)

In ∆ABD

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

AB/9 = 1/ tanθ

9/AB = tanθ         -(2)

From (1) and (2)

AB/4 = 9/AB

AB2 = 36

AB = √36

AB = √(6 * 6)  

AB = 6m

Height of the tower is 6m.



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