Observing from the top of a tower,the angles | vedclass

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(B) Let $H$ be the height of the tower and $h = 100 \, m$ be the height of the building.Let $x$ be the distance between the tower and the building.Fro

Vedclass · Accepted Answer

$150$

Vedclass · Answer

(B) Let $H$ be the height of the tower and $h = 100 \, m$ be the height of the building.Let $x$ be the distance between the tower and the building.From the top of the tower,the angle of depression to the top of the building is $30^{\circ}$,so the angle of elevation from the top of the building to the top of the tower is $30^{\circ}$.Thus,$	an(30^{\circ}) = \frac{H - 100}{x} \implies \frac{1}{\sqrt{3}} = \frac{H - 100}{x} \implies x = \sqrt{3}(H - 100)$.From the top of the tower,the angle of depression to the bottom of the building is $60^{\circ}$,so the angle of elevation from the bottom of the building to the top of the tower is $60^{\circ}$.Thus,$	an(60^{\circ}) = \frac{H}{x} \implies \sqrt{3} = \frac{H}{x} \implies x = \frac{H}{\sqrt{3}}$.Equating the two expressions for $x$: $\sqrt{3}(H - 100) = \frac{H}{\sqrt{3}}$.$3(H - 100) = H \implies 3H - 300 = H \implies 2H = 300 \implies H = 150 \, m$.

Observing from the top of a tower,the angles of depression of the top and the bottom of a $100 \, m$ high building are found to be $30^{\circ}$ and $60^{\circ}$ respectively. Find the height of the tower in $m$.

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