Question

Two poles of different heights are erected from the ground. If the smaller pole is $20m$ high and distance between two poles are $10m,$ and the angle of elevation of the top of the longer pole from that of the shorter pole is ${60}^{\circ }$, find the length of the second pole.

• A. $20+\frac{10}{\sqrt{3}}m$
• B. $20+5\sqrt{3}m$
  
• C. $10+10\sqrt{3}m$
• D. $20+10\sqrt{3}m$

Answer 1

The correct option is D

$20+10\sqrt{3}m$

Distance between the two poles, $AC=10m$
Height of the shorter pole, $AE=CD=20m$

Let $BC=xm$
Consider the right angled triangle ABC,
$\tan {60}^{\circ }=\frac{\text{opposite side}}{\text{adjacent side}}=\frac{BC}{AC}=\frac{x}{10}$
$\Rightarrow x=10\times \tan {60}^{\circ }$
$\because \tan {60}^{\circ }=\sqrt{3}$
$\therefore x=10\sqrt{3}m$
Thus, height of second pole $=CD+BC=20+10\sqrt{3}m$