The angle of depression of the base of a towe | vedclass

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(C) Let the height of the cliff be $H$ and the distance between the cliff and the tower be $d$. Let the height of the tower be $h = 50 \, m$.From the

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$150$

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(C) Let the height of the cliff be $H$ and the distance between the cliff and the tower be $d$. Let the height of the tower be $h = 50 \, m$.From the base of the cliff,the angle of elevation to the top of the tower is $30^{\circ}$. Thus,$	an(30^{\circ}) = \frac{h}{d} \implies \frac{1}{\sqrt{3}} = \frac{50}{d} \implies d = 50\sqrt{3} \, m$.From the top of the cliff,the angle of depression to the base of the tower is $60^{\circ}$. Thus,$	an(60^{\circ}) = \frac{H}{d} \implies \sqrt{3} = \frac{H}{50\sqrt{3}}$.Solving for $H$,we get $H = 50\sqrt{3} 	imes \sqrt{3} = 50 	imes 3 = 150 \, m$.

The angle of depression of the base of a tower from the top of a cliff is $60^{\circ}$ and the angle of elevation of the top of the tower from the base of the cliff is $30^{\circ}$. If the height of the tower is $50 \, m$,find the height of the cliff in meters.

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