Question

Two poles stand at either side of the road
Height of the poles are in the ratio 1:2
The road is 90 feet wide
The angle of elevation of both the poles from a point in between the road is 60degree find the height of the poles

Answer 1

$Let \: AB \:and \:CD \:are \:two \:poles .$

$AB : CD = 1 : 2 \: ( given )$

$Let \: AB = x \:ft , \: CD = 2x \:ft$

$"O" \:is \:the \:point \:from \:where \:the \\elevation \:angles \:are \:measured .$

$Distance \: between \:two \:poles (BD) = 90 \:ft$

$Let \: BO = y \:ft , \\and \: OD = (90 - y) \:ft$

$In \: \triangle AOB , we \:have ,\\tan \: \angle {AOB} = \frac{AB}{BO}$

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

$\implies \sqrt{3} = \frac{x}{y}$

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

$In \: \triangle COD , we \:have ,\\tan \: \angle {COD} = \frac{CD}{OD}$

$\implies tan 60 \degree = \frac{2x}{(90-y)}$

$\implies \sqrt{3} = \frac{2x}{(90-y)}$

$\implies 90 - y = \frac{2x}{\sqrt{3}}$

$\implies y = 90 - \frac{2x}{\sqrt{3}}\: ---(2)$

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

$\frac{x}{\sqrt{3}} = 90 - \frac{2x}{\sqrt{3}}$

$\implies \frac{x}{\sqrt{3}} + \frac{2x}{\sqrt{3}} = 90$

$\implies \frac{3x}{\sqrt{3}} =90$

$\implies x = 90 \times \frac{\sqrt{3}}{3} \\= 30\sqrt{3}$

$Now, Height \:of \:the \:pole (AB) = 30\sqrt{3} \:ft$

$Now, Height \:of \:the \:pole (CD) \\= 2x\\= 2 \times 30\sqrt{3} \\= 60\sqrt{3} \:ft$

•••♪