For a man,the angle of elevation of the highe | vedclass

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(A) Let the height of the temple be $h$ and the initial distance of the man from the base of the temple be $x$.From the first position,$	an(60^\circ)

Vedclass · Accepted Answer

$60\sqrt{6} \ m$

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(A) Let the height of the temple be $h$ and the initial distance of the man from the base of the temple be $x$.From the first position,$	an(60^\circ) = \frac{h}{x} \implies x = \frac{h}{\sqrt{3}}$.After walking $240 \ m$ north,the man is at a distance $d = \sqrt{x^2 + 240^2}$ from the base of the temple.From the second position,$	an(30^\circ) = \frac{h}{d} \implies d = h\sqrt{3}$.Substituting $d$ and $x$ in the distance formula: $(h\sqrt{3})^2 = (\frac{h}{\sqrt{3}})^2 + 240^2$.$3h^2 = \frac{h^2}{3} + 57600$.$3h^2 - \frac{h^2}{3} = 57600 \implies \frac{8h^2}{3} = 57600$.$h^2 = \frac{57600 	imes 3}{8} = 7200 	imes 3 = 21600$.$h = \sqrt{21600} = \sqrt{3600 	imes 6} = 60\sqrt{6} \ m$.

For a man,the angle of elevation of the highest point of the temple situated east of him is $60^\circ$. On walking $240 \ m$ to the north,the angle of elevation is reduced to $30^\circ$. The height of the temple is:

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