Question

From a point on the ground 40 m away from the foot of a tower, the angle of elevation of the top of the tower is 30°. The angle of elevation of the top of a water tank (on the top of the tower) is 45°.
Find :
(i). Height of the tower
(ii). Depth of the tank​

Answer 1

Let the height of the tower above the foot of the tower = h metre

Angle of elevation of the tower top from a point 40 m away from the foot of tower = 30°

==> h/40 = Tan 30° = 1/√3. Therefore height of the tower = 40 × 1/√3 = 40/√3 m= 23. 094 m ~ 23.1 m

Let the combined height of the tower and the water tank atop it from the foot of the tower = H

==> H/ 40 = Tan 45° or H = 40× Tan45° = 40m.

Therefore height of water tank above the top if the tower = 40m - 23.1 m = 16.9 m.

A water tank of 16.9 m height is way too high

Answer 2

To find :-

• Height of the tower = ?

• Depth of the tank = ?

Given :-

• Distance from the ground = 40m

• The angle of elevation at the top of the tower = 30°

• The angle of elevation at the top of the water tank = 45°

Solution :-

Let the height of the tower be considered as - h

.•. h/40 = tan30°

=> h/40 = 1/√3 [ value of tan 30° = 1/√3 ]

=> h = 40/√3

=> h = 40/1.732 [√3 = 1.732 ]

=> h = 23.094 m

Let the total height from the tower's foot be considered as - h'

.•. h'/40 = tan45°

=> h'/40 = 1 [ tan45° = 1 ]

=> h' = 40m

Thus, the required height = 40m - 23.094m = 16.906m