The angle of elevation of the top of a vertic | vedclass

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(B) Let the height of the tower be $h$ and the distance between the tower and point $P$ be $d$.From point $P$,the angle of elevation to the top of the

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(B) Let the height of the tower be $h$ and the distance between the tower and point $P$ be $d$.From point $P$,the angle of elevation to the top of the tower is $60^{\circ}$,so $	an 60^{\circ} = h / d$,which implies $d = h / \sqrt{3}$.Point $Q$ is $10\, m$ vertically above $P$. The angle of depression to the foot of the tower is $30^{\circ}$. This forms a right-angled triangle with the height $10\, m$ and base $d$,so $	an 30^{\circ} = 10 / d$,which implies $d = 10 / 	an 30^{\circ} = 10\sqrt{3}$.Equating the two expressions for $d$: $h / \sqrt{3} = 10\sqrt{3}$.Therefore,$h = 10\sqrt{3} 	imes \sqrt{3} = 10 	imes 3 = 30\, m$.

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