Page No 141
Q1: 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).
Ans: Given: length of the rope (AC) = 20 m, ∠ACB = 30°
Let the height of the pole (AB) = h metre
⇒ h/20 = 1/2⇒ h = 20/2 = 10 m
Hence, height of the pole = 10 m
Q2: A tree breaks due to a 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.
Ans: Let DB is a tree and AD is the broken part of it that touches the ground at C.
Given: ∠ACB = 30º
and BC = 8m
Let AB = x m
and AD = y m
∴ Now, length of the tree
= (x + y) m
In Δ ABCHence, total height of the tree =
Q3: 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 older 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?
Ans: Let l1 is the length of the slide for children below the age of 5 years and l2 is the length of the slide for elder children.
In ΔABC
Q4: 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.
Ans: Let h be the height of the tower
⇒
⇒
Q5: 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.
Ans: Given: height AB = 60 m, ∠ACB = 60°, AC = length of the string
Hence, the length of the string = 40√3 m
Q6: 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.
Ans: Let AB = height of the building
The distance walked by the boy towards building
DE = DF – EF
Q7: 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.
Ans: Given: AB = 20 m (Height of the building)
Let AD = h m (Height of the tower)
Hence, height of the tower =
Page No 142
Q8: 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.
Ans: Let the height of the pedestal AB = h m
Given: height of the statue = 1.6 m, ∠ACB = 45° and ∠DCB = 60°
Hence, height of the pedestal
Q9: 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.
Ans: Given: height of the tower AB = 50 m
∠ACB = 60°, ∠DBC = 30°
Let the height of the building
CD = x m
Q10: 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.
Ans: Let AB = CD = h m [Height of the poles]
Given: BC = 80 m [Width of the road]
Let CE = x m
∴ BE = (80 – x) m
In ΔCDE,
In ΔABE,
… (ii)
From equation (i) and (ii), we get
Substituting h in equation (i),
80 – x = 20 m
Hence, position of the point is at a distance of 60 m from pole CD and 20 m from pole AB.
Q11: 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.
Ans: Let the height of the tower AB = h m and BC be the width of the canal.
Given: ∠ACB = 60° and ∠ADB = 30°
⇒
⇒
⇒
Hence, the height of the tower = 10√3 m and width of the canal = 10 m.
Q12: 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.
Ans: Let height of the tower AB = (h + 7) m
Given: CD = 7m (height of the building),
∠ACE = 60°, and ∠ECB = 45°
⇒ ∠CBD = 45º
⇒
⇒
Q13: 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.
Ans: Given: height of the lighthouse = 75 m
Let C and D are the positions of two ships.
We have ∠XAD = ∠ADB = 30°
and ∠XAC = ∠ACB = 45°
⇒
Hence, the distance between two ships is 54.75 m.
Q14: 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° (see figure). Find the distance travelled by the balloon during the interval.
Ans: Let the first position of the balloon is A and after some time it will reach to the point D. The vertical height ED = AB = (88.2 – 1.2) m = 87 m.
Distance travelled by the balloon from A to D is BE.
So, BE = CE – CB
Q15: 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.
Ans: Let the height of the tower AB = h m
Given: ∠XAD = ∠ADB = 30°
and ∠XAC – ∠ACB = 60°
Let the speed of the car = x m/sec