Two poles stand at either side of the roadHeight of the poles are in t

1.	Two poles stand at either side of the roadHeight of the poles are in the ratio 1:2The road is 90 feet wideThe angle of elevation of both the poles from a point in between the road is 60degree find the height of the poles
Answer» $Let \: AB \:and \:CD \:are \:two \:poles .$ $AB : CD = <klux>1</klux> : <klux>2</klux> \: ( given )$ $Let \: AB = x \:ft , \: CD = 2x \:ft$ $\text{[math]}$ $Distance \: between \:two \:poles (BD) = 90 \:ft$ $Let \: BO = y \:ft , \\and \: OD = (90 - y) \:ft$ $In \: \triangle AOB , we \:have ,\\tan \: \angle {AOB} = \frac{AB}{BO}$ $\implies tan 60 \degree = \frac{x}{y}$ $\implies \sqrt{3} = \frac{x}{y}$ $\implies y = \frac{x}{\sqrt{3}} \: ---(1)$ $In \: \triangle COD , we \:have ,\\tan \: \angle {COD} = \frac{CD}{OD}$ $\implies tan 60 \degree = \frac{2x}{(90-y)}$ $\implies \sqrt{3} = \frac{2x}{(90-y)}$ $\implies 90 - y = \frac{2x}{\sqrt{3}}$ $\implies y = 90 - \frac{2x}{\sqrt{3}}\: ---(2)$ /* From (1) and (2) */ $\frac{x}{\sqrt{3}} = 90 - \frac{2x}{\sqrt{3}}$ $\implies \frac{x}{\sqrt{3}} + \frac{2x}{\sqrt{3}} = 90$ $\implies \frac{3x}{\sqrt{3}} =90$ $\implies x = 90 \times \frac{\sqrt{3}}{3} \\= 30\sqrt{3}$ $Now, Height \:of \:the \:pole (AB) = 30\sqrt{3} \:ft$ $Now, Height \:of \:the \:pole (CD) \\= 2x\\= 2 \times 30\sqrt{3} \\= 60\sqrt{3} \:ft$ •••♪