Question

A 1.4m tall boy stands at a point 50m away from a tower and observes the angle of elevation of the top of the tower to be 60°. Find the height of the tower. 

Answer 1

Let the position of the boy be at point T. 
BR = TQ = 50 m 
RQ = BT= 1.5 m

In ΔPRB 

`"PR"/"BR" = tan 60^circ`

`"PR"/50 = sqrt(3)`

`"PR" = 50sqrt(3)`

Height of the tower 

`= "PQ" = "PR + RQ" = 50sqrt(3) + 1.5 = 50 xx 1.732 + 1.5 = 86.6 + 1.5 = 88.1  "m"`

Thus , the height of the tower is approximately 88 m.