Watching from the top of an 80 , m high hill, | vedclass

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Question

(B) Let the height of the hill be $h_1 = 80 \, m$ and the height of the tower be $H = h_1 + h_2$,where $h_2$ is the height of the tower above the leve

Vedclass · Accepted Answer

$80(1 + \frac{1}{\sqrt{3}}) \, m$

Vedclass · Answer

(B) Let the height of the hill be $h_1 = 80 \, m$ and the height of the tower be $H = h_1 + h_2$,where $h_2$ is the height of the tower above the level of the hill top.Let $d$ be the horizontal distance between the hill and the tower.From the angle of depression of the base: $	an(45^\circ) = \frac{h_1}{d} \implies 1 = \frac{80}{d} \implies d = 80 \, m$.From the angle of elevation of the top: $	an(30^\circ) = \frac{h_2}{d} \implies \frac{1}{\sqrt{3}} = \frac{h_2}{80} \implies h_2 = \frac{80}{\sqrt{3}} \approx 46.19 \, m$.The total height of the tower is $H = 80 + 46.19 = 126.19 \, m$ (or $80(1 + \frac{1}{\sqrt{3}}) \, m$).

Watching from the top of an $80 \, m$ high hill,the angle of elevation of the top of a tower is $30^\circ$ and the angle of depression of the base of the tower is $45^\circ$. Find the height of the tower in $m$.

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