Question

A boy $1.7$ m tall, is $25$ m away from a tower and observes the angle of elevation of the top of the tower to be ${60}^o$. Find the height of the tower.

• A. $55$ m
• B. $45$ m
• C. $20$ m
• D. $35$ m

Answer 1

The correct option is B $45$ m

Let $A{A}^{\prime }$ be the boy and the tower be $B^{\prime }C$
Therefore, $A{A}^{\prime }=1.7$ m $=B{B}^{\prime }$
and $A^{\prime }{B}^{\prime }=25m=AB$
Now, angle of elevation is $\angle CAB=\theta$.
Given, $\theta ={60}^o$
$\Rightarrow \tan \theta =\frac{BC}{AB}$
$\Rightarrow 25\tan {60}^o=BC$
$\Rightarrow BC=25\sqrt{3}$
Now, height of tower
$=BC+B{B}^{\prime }$
$=25\sqrt{3}+1.7$
$=45$ m