Question

Two poles of different heights are erected from the ground. If the smaller pole is 20 m high and distance between the two poles is 10 m, and the angle of elevation of the tip 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 $20+10\sqrt{3}m$

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

Let BC = $xm$

Consider the right angled triangle ABC. Then,

$tan{60}^{\circ }=\frac{oppositeside}{adjacentside}=\frac{BC}{AC}=\frac{x}{10}$

$x=10\times tan{60}^{\circ }(tan{60}^{\circ }=\sqrt{3})$

$\Rightarrow$ $x=10\sqrt{3}m$

$\therefore$ Height of the second pole = $CD+BC=20+10\sqrt{3}m$