# Height and Distance - Aptitude Questions and Answers - RejinpaulPlacement

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:

Solution:   Let AB be the tower. Then, ∠APB = 30° and AB = 100 m

AB/x = tan 30° = 1/√3

=> AP = (AB * √3 ) = 100√3 m

AP = (100 * 1.73) m = 173 m
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:

Solution:   Let AB be the wall and BC be the ladder.

Then, ∠ACB = 60° and AC = 4.6 m

AC/BC = cos 60° = 1/2

=> BC = 2 * AC = (2 * 4.6) m = 9.2 m

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:

Solution:   Let AB be the observer and CD be the tower.

Draw BE CD

Then, CE = AB = 1.6 m, BE = AC = 20√3

DE/BE = tan 30° = 1/√3

=> DE = (20 √3)/√3 m = 20 m

CD = CE + DE = (1.6 + 20) m = 21.6 m

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:

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 ?

Solution:   One of AB, AD and CD must have been given.

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:

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
=> 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?

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