Question

Two poles of equal heights are standing opposite to each other on either side of the road, which is 120 feet wide. From a point between them on the road, the angles of elevation of the top of the poles are 60º and 30º respectively. Find the height of the poles and the distances of the point from the poles.

Answer 1

Answer to your question

Answer 2

$Let \: height \:of \:the \:equal \:poles \\AB = CD = h \:feet$

$Distance \: between \:two \:poles (BC) = 120\ft$

$Point \:of \: observation \: is \: E$

$Let \: BE = x \:ft$

$EC = ( 120 - x ) \:ft$

$i ) In \: triangle ABE , \\tan \angle {AEB} = \frac{AB}{BE}$

$\implies tan 60\degree = \frac{h}{x}$

$\implies \sqrt{3} x = h \: ---(1)$

$ii )In \: triangle DEC , \\tan \angle {DEC} = \frac{DC}{EC}$

$\implies tan 30\degree = \frac{h}{(120-x)}$

$\implies \frac{1}{\sqrt{3} } (120 - x ) = h \: ---(2)$

/* From (1) and (2) */

$\sqrt{3} x = \frac{1}{\sqrt{3} } (120 - x )$

$\implies \sqrt{3}^{2} x = 120 - x$

$\implies 3x = 120 - x$

$\implies 4x = 120$

$\implies x = \frac{120}{4} \\= 30\:ft\: --(3)$

/* put value of x in equation (1) , we get */

$h = \sqrt{3} \times 30$

$\implies h =30 \sqrt{3}$

$\implies h =39 \times 1.732$

$\implies h = 51.96 \:ft$

Therefore.,

$\red{ Height \:two \:equal \:poles } \green {= 51.96\:ft }$

$\red{ Distance \:from \:B \: to \:the \:point} \green {= 30 \:ft }$

$\red{ Distance \:from \:C \: to \:the \:point} \\=(120 - x)\\= (120-30) \\\green {= 90 \:ft }$

•••♪