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Question

(B) Let $PQ$ represent the pillar with height $h$.In $\Delta PAQ$,$	an(30^o) = \frac{PQ}{AQ}$ $\Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{AQ}$ $\Right

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) Let $PQ$ represent the pillar with height $h$.In $\Delta PAQ$,$	an(30^o) = \frac{PQ}{AQ}$ $\Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{AQ}$ $\Rightarrow AQ = h\sqrt{3}$.In $\Delta PQB$,$	an(60^o) = \frac{PQ}{BQ}$ $\Rightarrow \sqrt{3} = \frac{h}{BQ}$ $\Rightarrow BQ = \frac{h}{\sqrt{3}}$.The distance covered in $10 	ext{ minutes}$ is $AB = AQ - BQ = h\sqrt{3} - \frac{h}{\sqrt{3}} = \frac{3h - h}{\sqrt{3}} = \frac{2h}{\sqrt{3}}$.Since the speed is uniform,the time taken is proportional to the distance.Time taken to cover $AB = 10 	ext{ minutes}$.Time taken to cover $BQ = \frac{BQ}{AB} 	imes 10 = \frac{h/\sqrt{3}}{2h/\sqrt{3}} 	imes 10 = \frac{1}{2} 	imes 10 = 5 	ext{ minutes}$.

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