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

Let BC be the tower and CD be the water tank.
We have,
AB  = 40m, ∠BAC = 30°  and ∠BAD =  45°
In ΔABD,

` tan 45° = (BD)/(AB)`

`⇒ 1 = (BD)/40`

⇒ BD = 40 m

Now, in ΔABC,

`tan 30° = (BC)/(AB)`

`⇒1/ sqrt(3) = (BC) /40`

`⇒ BC = 40/sqrt(3)`

`⇒ BC = 40/ sqrt(3) xx sqrt(3)/sqrt(3)`

`⇒ BC = (40sqrt(3) )/3 m`

`"(i) The height of the tower," BC = (40 sqrt(3))/3 = (40 xx 1.73)/3 = 23.067  ~~ 23.1 m`

`"(ii) The depth of the tank " CD = (BD - BC) = (40-23.1 ) = 16.9 m`