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) the height of the tower, (ii) the depth of the tank.

Answer 1

Draw a figure based on given instructions:

From figure: A be the point of observer

BC = Tower

CD = Water tank

∠BAD = 45 degrees, ∠BAC = 30 degrees and AB = 40 m

From right triangle ABD,

tan 45° = BD/AB

1 = BD/40

BD = 40 …….(1)

From another right triangle, ABC

Tan 30° = BC/AB

1/ √3 = BC/40

Or BC = 40/√3

Or BC = 40 √3/ 3 …(2)

Now,

(i) Height of the tower = BC = 40 √3/3 = 23.1

Height of the tower is 23.1 m.

(ii) Depth of the tank = CD = BD – BC = 40-23.1 = 16.9

Therefore, depth of the tank is 16.9 m.