


8.
From a point P on a level ground, the angle of elevation of the top of a tower is 30°. If the tower is 100 m high, the distance of point P from the foot of the tower is:
9.
The angle of elevation of a ladder leaning against a wall is 60° and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is:
10.
An observer 1.6 m tall is 20√3 m away from a tower. The angle of elevation from the eye to the top of the tower is 30°. The height of the tower is:
11.
Two ships are sailing in the sea on the two sides of a lighthouse. The angles of elevation of the top of the lighthouse as observed from the two ships are 30° and 45° respectively. If the lighthouse is 100 m high, the distance between the two ships is:
Answer & Solution
Answer: (C) 273 m
Solution: Let AB be the lighthouse and C and D be the positions of the ships. Then, AB = 100 m, ∠ACB = 30° and ∠ADB = 45° AB/AC = tan 30° = 1/√3 =>AC = AB * √3 = 100√3 m AB/AD = tan 45° = 1 => AD = AB = 100 m CD = (AC + AD) = (100√3 + 100) m CD = 100 (√3 + 1) m = (100 * 2.73) m = 273 m 12.
A man standing at a point P is watching the top of a tower, which makes an angle of elevation of 30° with the man’s eye. The man walks some distance towards the tower to watch its top and the angle of elevation becomes 60°. What is the distance between the base of the tower and the point P ?
13.
The angle of elevation of the top of a tower from a certain point is 30°. If the observer moves 20 m towards the tower, the angle of elevation of the top of the tower increases by 15°. The height of the tower is:
Answer & Solution
Answer: (C)27.3 m
Solution: Let AB be the tower and C and D be the points of observation. Then, ∠ACB = 30°, ∠ADB = 45° and Cd = 20 m Let AB = h Then, AB/AC = tan 30° = 1/√3 => AC = AB * √3 = h√3 And, AB/AD tan 45° = 1 => AD = AB = h CD = 20 (AC  AD) = 20 => h√3  h = 20 h = 20/(( √3  1)) * (√3 + 1)/(√3 + 1) = 10 (√3 + 1) m = (10 * 2.73) m = 27.3 m 14.
A man is watching from the top of a tower a boat speeding away from the tower. The boat makes an angle of depression of 45° with the man’s eye when at a distance of 60 metres from the tower. After 5 seconds, the angle of depression becomes 30°. What is the approximate spped of the boat, assuming that it is running in still water?
Answer & Solution
Answer: (B)32 kmph
Solution: Let AB be the tower and C and D be the two positions of the boats. Then, ∠ACB = 45°, ∠ADB = 30° and AC = 60 m Let AB = h Then, AB/AC = tan 45° = 1 => AB = AC => h = 60 m And, AB/AD = tan 30° = 1/√3 => AD = (AB * √3 ) = 60√3 m CD = (AD  AC) = 60 (√3  1) m Hence, required speed = ( (60 (√3  1) )/5 ) m/s = (12 * 0.73) m/s \ = (12 * 0.73 * 18/5 ) km/hr = 31.5 km/hr = 32 km/hr 