Question

The angle of elevation of the top of a tower from a point $A$ on the ground is ${30}^o$. On moving a distance of $30$m towards the foot of the tower to a point $B$, the angle of elevation increases to ${45}^o$. The distance of the tower from point $A$ is

• A. $41$m
• B. $71$m
• C. $63$m
• D. None

Answer 1

The correct option is B $71$m

Let height of tower is h and distence of foot of tower from point B is x

In $\Delta$PBQ ,$\angle PBQ={45}^o$

$\therefore tan45=\frac{h}{x}$

$\Rightarrow 1=\frac{h}{x}$

$\Rightarrow x=h$..........................(1)

In $\Delta$PAQ ,$\angle PAQ={30}^o$

$\therefore tan{30}^o=\frac{h}{x+30}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x+30}$

Put the value of x=h as per(1)

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{x}{30+x}$

$\sqrt{3}x=30+x\Rightarrow 1.732x-x=30\Rightarrow x=\frac{30}{0.732}\Rightarrow x=40.983\approx 41$

Then distence of tower from point $A=41+30=71m$