Question

An observer 1.5 m tall is 28.5 m away from a tower and the angle of elevation of the top of the tower from the eye of the observer is ${45}^{\circ }$. The height of the tower is

(a) 27 m

(b) 30 m

(c) 28.5 m

(d) none of these

Answer 1

Given the height of the observer be DE = 1.5 m
That is AB = 1.5 m
Let BC = h is the height of the tower
Hence AC = (h – 1.5) m
Given the distance between the observer and the tower is AD = BE = 28.5 m
In right $△CAD,\theta ={45}^o$

$tan{45}^o=\frac{AC}{AD}\Rightarrow 1=\frac{h-1.5}{28.5}\Rightarrow 28.5=h-1.5\Rightarrow h=28.5+1.5-30m$

Thus the height of the tower is 30 m.