These NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1 Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 10 Maths 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:
In this fig. AB is height of the pole and AC is length of the rope. The inclination of rope with the ground is 30°.
Now, In ∆ABC
sin 30° = \(\frac { AB }{ AC }\)
or, \(\frac { 1 }{ 2 }\) = \(\frac { AB }{ 20 }\)
AB = \(\frac { 20 }{ 2 }\) = 10m
Therefore, the height of the pole is AB = 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:
Let AB be the tree broken at a point C such that the broken part CB takes the position CO and strikes the ground at O. It is given that OA = 8 metres and ∠AOC = 30°.
Height of the true = (x + y)m
Hence height of the tree is 8\(\sqrt{3}\) 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:
I – case (The children below 1 the age of 5 years)
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 A and B be the top and foot of the tower AB, Let BC be the horizontal ground. It is given that BC = 30m, ∠ACB = 30° and ∠B = 90°
Let x be the height of the tower AB, i.e., AB = x m
We have cos 30° = \(\frac { BC }{ AC }\)
Hence, height of the tower is 10\(\sqrt{3}\) 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:
In this fig. AB is the height of kite and AB is length of string which is tied at point C on the i ground. At point C the inclination of the string with the ground is 60°.
Now, In ∆ABC,
Therefore, the length of string = 40\(\sqrt{3}\) 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:
In this fig, AB is height of the building and CD is the height of the boy. Angle of elevation from boy’s eyes to the top of the building is 30° and after y m walk’s towards building the angle of elevation becomes 60°.
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 this fig. AB is height of tower which is kept on the to a 20m high tower BC Angle of elevation of the top and bottom of tower from a 1 point D on ground is 60° and 45° repectivelv.
Hence, the height of the tower is 20(\(\sqrt{3}\) – 1) 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:
In this fig. AB is the height of the statue and BC is the height of pedestal. At the poind D on the ground the angle of elevation of the top of the statue and top of pedestal is 60° and 45° respectively.
Therefore, the height of pedestal is BC = 2.19 m (approx)
Question 9.
The angle of elevation of the top of a building from the foot of a 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 this fig, ABC is the height of tower and CD is the height of building. Angle of elevation of the top of the building from the foot of a tower is 30° and the angle of elevation of the top of the tower from the foot of a building is 60°.
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 distance of the point from the poles.
Solution:
In this fig, AB and CD are poles of equal height on either side of 80 m wide road. There is a point P on the road in which the angle of elevation of the poles are 60° and 30° respectively.
Let the distance between the point P and the first pole is x m
∴ The distance between the point P and the second pole is 80 – x m
Therefore, height of each poles is 20\(\sqrt{3}\) m and distance of the point from first pole is x = 20m and (80 – x) = (80 – 20) = 60m
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 the given figure). Find the height of the tower and the width of the CD and 20 m from pole AB.
Solution:
In this fig, AB is the height of the cable tower and CD is the height of the building. The angle of elevation of the top of a cable tower from the top of the building is 60° and angle of depression of the foot of a cable tower from the top of the building is 45°
In ∆ABC,
So, height of the three = AB = 10\(\sqrt{3}\) m and width of the river = BC = 10m
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 this fig. AB is the height of the cable towar and CD is the height of the building. The angle of elevation of the top of a cable towar from the top of the building is 60 and the angle of depression of the foot of a cable tower from the top of tire building is 45°.
Therefore, height of the tower AB = AE + EB = 7\(\sqrt{3}\) + 7 = 7(\(\sqrt{3}\) + 1)
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 this fig, AB is the height of the lighthouse and at point C and D there are two ships just behind each other. Angle of depression from the top of the lighthouse to the ships are 30° and 45° respectively.
Therefore, the distance between the two ships are
CD = BD – BC
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 sometime, the angle of elevation reduces to 30° (see figure). Find the distance travelled by the balloon during the interval.
Solution:
Initial height = 88.2 – height of the girl
= 88.2 – 1.2 = 87 m
Therefore, the distance travelled by the ballon during the interval CQ \(\frac{296 \sqrt{3}}{5}\) 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 AB be a tower and a man be at A.
A man observes a car at D at an angle of depression of 30°
i.e., ∠EAD = 30°
⇒ ∠EAD = 30° [Both are alternate angels; AE || BD and AD cutsthen at A and D]
The car is approaching towards B with a uniform speed. After travelling for 6 seconds let the car be at C.
From A, the angle of depression of the car at C is 60°.
i.e., ∠EAC= 60°
⇒ ∠EAC = ∠ACB
[Both are alternate angles; AE || BD and AC cuts them at A and C]
In right triangle ABC, we have,
From equations (i) and (ii), we get
Let V be the velocity of the car
Now in 6 sec distance covered = CD
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:
Angles are complementary.
∠DAC = 9
and ∠DBC = 90° – 0
In ∆DBC,
Hence, height of tower is 6m Hence Proved.