From the top of a hill 100, m high,the angles | vedclass

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(A) Let the height of the hill be $H = 100\, m$. Let the height of the tower be $h$. Let the distance between the hill and the tower be $x$.From the g

Vedclass · Accepted Answer

$100(1 - \frac{1}{\sqrt{3}})$

Vedclass · Answer

(A) Let the height of the hill be $H = 100\, m$. Let the height of the tower be $h$. Let the distance between the hill and the tower be $x$.From the geometry of the problem,for the bottom of the tower (angle of depression $45^{\circ}$):$	an(45^{\circ}) = \frac{H}{x} \implies 1 = \frac{100}{x} \implies x = 100\, m$.For the top of the tower (angle of depression $30^{\circ}$),the height of the hill above the level of the tower top is $H - h$:$	an(30^{\circ}) = \frac{H - h}{x} \implies \frac{1}{\sqrt{3}} = \frac{100 - h}{100}$.$100 - h = \frac{100}{\sqrt{3}} \implies h = 100 - \frac{100}{\sqrt{3}} = 100(1 - \frac{1}{\sqrt{3}})$.Using $\sqrt{3} \approx 1.732$,$h = 100(1 - 0.577) = 100(0.423) = 42.3\, m$ (approx).

From the top of a hill $100\, m$ high,the angles of depression of the top and the bottom of a tower are observed to be $30^{\circ}$ and $45^{\circ}$ respectively. Find the height of the tower in $m$.

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