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Question

# The angle of elevation of the top of a vertical tower from a point on the ground is 60°. From another point 10 m vertically above the first, its angle of elevation is 30°. Find the height of the tower. [CBSE 2011]

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Solution

## Let PQ be the tower. We have, $\mathrm{AB}=10\mathrm{m},\angle \mathrm{MAP}=30°\mathrm{and}\angle \mathrm{PBQ}=60°\phantom{\rule{0ex}{0ex}}\mathrm{Also},\mathrm{MQ}=\mathrm{AB}=10\mathrm{m}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Let}\mathrm{BQ}=x\mathrm{and}\mathrm{PQ}=h\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{AM}=\mathrm{BQ}=x\mathrm{and}\mathrm{PM}=\mathrm{PQ}-\mathrm{MQ}=h-10\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{In}∆\mathrm{BPQ},\phantom{\rule{0ex}{0ex}}\mathrm{tan}60°=\frac{\mathrm{PQ}}{\mathrm{BQ}}\phantom{\rule{0ex}{0ex}}⇒\sqrt{3}=\frac{h}{x}\phantom{\rule{0ex}{0ex}}⇒x=\frac{h}{\sqrt{3}}.....\left(\mathrm{i}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\mathrm{in}∆\mathrm{AMP},\phantom{\rule{0ex}{0ex}}\mathrm{tan}30°=\frac{\mathrm{PM}}{\mathrm{AM}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{\sqrt{3}}=\frac{h-10}{x}\phantom{\rule{0ex}{0ex}}⇒h\sqrt{3}-10\sqrt{3}=x\phantom{\rule{0ex}{0ex}}⇒h\sqrt{3}-10\sqrt{3}=\frac{h}{\sqrt{3}}\left[\mathrm{Using}\left(\mathrm{i}\right)\right]\phantom{\rule{0ex}{0ex}}⇒3h-30=h\phantom{\rule{0ex}{0ex}}⇒3h-h=30\phantom{\rule{0ex}{0ex}}⇒2h=30\phantom{\rule{0ex}{0ex}}⇒h=\frac{30}{2}\phantom{\rule{0ex}{0ex}}\therefore h=15\mathrm{m}$ So, the height of the tower is 15 m.

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