A tower 50 m high stands on the top of a mou | vedclass

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(C) Let the height of the mount be $h$ and the distance from the point on the ground to the base of the mount be $d$. From the triangle formed by the

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$25(\sqrt{3} + 1) \ m$

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(C) Let the height of the mount be $h$ and the distance from the point on the ground to the base of the mount be $d$. From the triangle formed by the mount,we have $	an(60^o) = \frac{h}{d}$,so $d = \frac{h}{	an(60^o)} = \frac{h}{\sqrt{3}}$. From the triangle formed by the tower and the mount,the total height is $h + 50$ and the angle of elevation is $75^o$. Thus,$	an(75^o) = \frac{h + 50}{d}$. We know $	an(75^o) = 	an(45^o + 30^o) = \frac{	an(45^o) + 	an(30^o)}{1 - 	an(45^o)	an(30^o)} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} = \frac{(\sqrt{3} + 1)^2}{3 - 1} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$. Substituting $d = \frac{h}{\sqrt{3}}$ into the equation: $2 + \sqrt{3} = \frac{h + 50}{h/\sqrt{3}} = \frac{\sqrt{3}(h + 50)}{h}$. $h(2 + \sqrt{3}) = \sqrt{3}h + 50\sqrt{3}$. $2h + \sqrt{3}h = \sqrt{3}h + 50\sqrt{3}$. $2h = 50\sqrt{3} \Rightarrow h = 25\sqrt{3} \ m$.

$A$ tower $50 \ m$ high stands on the top of a mount. From a point on the ground,the angles of elevation of the top and bottom of the tower are found to be $75^o$ and $60^o$ respectively. The height of the mount is:

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