Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1 Textbook Exercise Questions and Answers.

## RBSE Class 10 Maths Solutions Chapter 9 Some Applications of Trigonometry Exercise 9.1

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 Figure)

Solution:

Let AB the height of the pole. AC = 20 m is the length of the rope.

In this situation, the angle of elevation is 30°. Now in right △ABC.

\(\frac{\mathrm{Perpendicular}}{\mathrm{Hypotenuse}}\) = \(\frac{\mathrm{AB}}{\mathrm{AC}}\) = sin 30°

[sin θ = \(\frac{\mathrm{Perpendicular}}{\mathrm{Hypotenuse}}\)

or \(\frac{\mathrm{AB}}{\mathrm{20}}\) = \(\frac{1}{2}\)

or AB = \(\frac{1}{2}\) × 20 = 10

Hence the height of the pole = 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

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:

Let the length of the tree before storm be BD. After storm AD = AC = length of the broken part of the tree.

Now in right △ABC,

Multiplying the numerator and the denominator by \(\sqrt {3}\)

Therefore height of the tree = 8\(\sqrt {3}\) m

or height of the tree = 8 × 1.732

= 13.856

≃ 13.86 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° 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:

Case First—For the children below the age of 5 years—

Let AC = l_{1}m be the length of the slide and BC = 1.5 m be the height of the slide. In this case angle of elevation is 30°.

In right of △ABC,

\(\frac{\mathrm{BC}}{\mathrm{AC}}\) = sin 30°

[∵ sin θ = \(\frac{perpendicular}{hypotenuse}\)]

or \(\frac{1.5}{l_{1}}\) = \(\frac{1}{2}\)

or l_{1} = 1.5 × 2 = 3 m

Case Second—For elder children—

Let AC = l_{2} m be the length of the slide and BC = 3 m be the height of the slide. In this case the angle of elevation is 60°.

In right △ABC,

multiplying the numerator and the denominator by \(\sqrt {3}\)

= \(\frac{6}{\sqrt{3}}\) × \(\frac{\sqrt{3}}{\sqrt{3}}\) = \(\frac{6 \sqrt{3}}{3}\) = 2\(\sqrt {3}\) m

Hence the length of the slide for the children below the age of 5 years = 3 m and for elder children the length of the slide = 2\(\sqrt {3}\) 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°. Find the height of the tower.

Solution:

Let QR = h m be the height of the tower and PQ = 30 m is the distance on the ground level.

In right △PQR,

= 10\(\sqrt {3}\) m = 10 × 1.732

h = 17.32 m (approximately)

Hence the height of the tower = 17.32 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°. Find the length of the string, assuming that there is no slack in the string.

Solution:

Let the position of the kite be at point R. PR = l m is the length of the string attached to the kite. In this case the angle of elevation is 60°.

or \(\frac{60}{l}\) = \(\frac{\sqrt{3}}{2}\) or \(\sqrt{3} l\) = 60 × 2

or l = \(\frac{60 \times 2}{\sqrt{3}}\) = \(\frac{120}{\sqrt{3}}\) \(\frac{\sqrt{3}}{\sqrt{3}}\)

multiplying the numerator and the denominator by \(\sqrt {3}\)

= \(\frac{120 \sqrt{3}}{3}\) = 40\(\sqrt {3}\)m

Hence, the length of the string = 40\(\sqrt {3}\) m

OR

Hence, the length of the string = 40 × 1.732

= 69.28 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 increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

Solution:

Let ED = 30 m be the height of the building and EC = 1.5 m be the height of the boy. In different cases the angles of elevation are respectively 30° and 60° and here DC = DE – EC = 30

– 1.5 = 28.5 m.

In right △ACD,

distance walked towards the building AB = AC – BC ,

= (x + y) – y

Hence, the distance walked towards the building = 19\(\sqrt {3}\)m

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 heigh building are 45° and 60° respectively. Find the height of the tower.

Solution:

Let BC = 20 m be the height of the building and DC = A m be the height of the transmission tower. The angles of elevation of the bottom and the top of a transmission tower fixed at the top of the building are 45° and 60° respectively.

or 20\(\sqrt {3}\) = 20 + h

or h = 20\(\sqrt {3}\) – 20

or h = 20(\(\sqrt {3}\) – 1) m

or = 20 (1.732 – 1) m

or = 20 × 0.732 = 14.64 m

Hence, height of the tower = 14.64 m

Hence the height of the transmission tower = 20(\(\sqrt {3}\) – 1) m = 14.64 m

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:

Let BC = h m be the height of the pedestal and CD = 1.6 m be the height of the statue. The angles of elevation of the top of the statue and the top of the pdedstal are 60° and 45° respectively.

Fron (i) and (ii),

= 0.8 × (\(\sqrt {3}\) + 1) m

Hence, height of the pedestal =0.8 × (\(\sqrt {3}\)+ 1) m

or height of the pedestal = 0.8 (1.732 + 1)

= 0.8 × 1.732 ≈ 2.20 m

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:

Let BC = 50 m be the height of the tower and AD = h m be the height of the building. The angle of elevation of the top of the building from the foot-point of the tower and the angle of elevation of the top of the tower from the foot-point of the building are 30° and 60° respectively.

Hence, height of the building 16\(\frac{2}{3}\) = 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:

Let BC = DE = h m be the heights of two equal poles and the point A be the required point from where the angles of elevation of two poles are 30° and 60°.

and in right △ABC,

or x = (80 – x) \(\sqrt {3}\) × \(\sqrt {3}\)

or x = (80 – x) 3

or x = 240 – 3x

or 4x = 240

or x = \(\frac{240}{4}\) = 60 m

substituting the value of x in (i)

h = \(\frac{60}{\sqrt{3}}\) = \(\frac{60}{\sqrt{3}}\) × \(\frac{\sqrt{3}}{\sqrt{3}}\)

(multiplying the numerator and the denominator by \(\sqrt {3}\))

= \(\frac{60 \sqrt{3}}{3}\) = 20\(\sqrt {3}\) m

Hence the height of the poles = (20 × 1.732) m = 34.64 m

∴ DA = x – 60 m

and AB = 80 – x = (80 – 60) m = 20 m

Hence, the height of pole is 34.64 m and the distances of the point from the poles are 20 m and 60 m respectively.

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

Solution:

Let BC = x m be the width of the canal and CD = h m be the height of the TV tower. In different cases the angles of elevation of the top of the tower are 30° and 60° respectively.

or \(\sqrt {3}\)(\(\sqrt {3}\)x) = 20 + x

or 3x = 20 + x

or 2x= 20

or x = \(\frac{20}{2}\) = 10

Substituting the value of x in (i)

h = 10 (\(\sqrt {3}\))

= 10 × 1.732 h

= 17.32 m

Hence the height of the TV tower is 17.32 m and the width of the canal is 10 m.

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:

Let BD = h m be the height of the cable tower and AE = 7 m be the height of the building. The angle of elevation of the top of the cable tower and th angle of depression of the foot are 60° and 45° respectively.

But AB = EC (Given)

7 = \(\frac{h-7}{\sqrt{3}}\) [Using (i) and (ii)]

or 7\(\sqrt {3}\) = h – 7

or h = 7\(\sqrt {3}\) + 7 = 7(\(\sqrt {3}\) + 1) m

Hence, height of the cable tower = 7(\(\sqrt {3}\) + 1) m

or h = 7(1.732 + 1) = 7(2.732)

or h= 19.124

or h = 19.124 m

Hence, height of the cable tower = 19.124 m

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:

Let CD = 75 m be the height of the light house and the angles of depression of two ships from the top point D of the light house are respectively 30° and 45°.

or \(\frac{x+y}{75}\) = \(\sqrt {3}\)

or x + y = 75\(\sqrt {3}\)

or x + 75 = 75\(\sqrt {3}\) [using (i)]

or x = 75\(\sqrt {3}\) – 75

or x = 75(\(\sqrt {3}\) – 1) m

or = 75(1.732 – 1)

or = 75(0.732)

or x = 54.90 m

Hence the distance between two ships = 75(\(\sqrt {3}\) – 1) m or 54.90 m

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 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° Fig. 9.13 (see Fig. 9.13). Find the distance travelled by the balloon during the interval.

Solution:

Let the position of 1.2 m tall girl be ‘A’. The angles of elevation of the balloon at different distances from this point are 30° and 60° respectively. Also, BE = CD = 87 m is the height of the balloon.

Hence, the distance covered by the balloon in this interval = 58 73 m

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:

Let CD = h m be the height of the tower and ‘A’ be the original position of the car and after 6 seconds the car reaches B. The angles of depression of the car at A and B are respectively 30° and 60°.

Again let the speed of the car be v metre per second.

∵ Distance = speed × time

AB = distance covered by car in 6 seconds.

AB = 6v m

and the time taken by car in reaching upto tower

= n seconds

∴ BC = nv m

In right △ACD,

Or 6v + nv = nv (\(\sqrt {3}\) × \(\sqrt {3}\))

Or 6v + nv = 3nv Or 6v = 2nv

Or n = \(\frac{6 v}{2 v}\) = 3

Hence the time taken by the car to reach the foot of the tower = 3 records.

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:

Let CD = h m be the height of the tower and B, A are the required points which are at distances of 4 m and 9 m respectively from the tower.

multiplying equations (i) and (ii)

Or \(\frac{1}{1}\) = \(\frac{h^{2}}{36}\)

∴ h^{2} = 36 Or h = 6 m

Hence the height of the tower = 6 m (Hence Proved)

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