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

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(C) Let $MN$ be the tower of height $h$ and $AN = x$ be the distance of the foot of the tower from point $A$.In $\Delta ANM$,$	an(45^o) = \frac{MN}{A

Vedclass · Accepted Answer

$15(3 + \sqrt{3})$

Vedclass · Answer

(C) Let $MN$ be the tower of height $h$ and $AN = x$ be the distance of the foot of the tower from point $A$.In $\Delta ANM$,$	an(45^o) = \frac{MN}{AN} = \frac{h}{x} = 1 \Rightarrow h = x$.Point $B$ is $30 \, m$ above $A$,so $PB = AN = x$ and $PM = MN - NP = h - 30 = x - 30$.In $\Delta BPM$,$	an(30^o) = \frac{PM}{PB} = \frac{x - 30}{x}$.$\frac{1}{\sqrt{3}} = \frac{x - 30}{x} \Rightarrow x = \sqrt{3}x - 30\sqrt{3}$.$x(\sqrt{3} - 1) = 30\sqrt{3} \Rightarrow x = \frac{30\sqrt{3}}{\sqrt{3} - 1}$.Rationalizing the denominator: $x = \frac{30\sqrt{3}(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{30(3 + \sqrt{3})}{3 - 1} = \frac{30(3 + \sqrt{3})}{2} = 15(3 + \sqrt{3}) \, m$.

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