A house subtends a right angle at the window | vedclass

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(B) Let the height of the first house be $H$. Let the window be at height $y$ from the ground. The distance between the houses is $d = 6 \ m$.From the

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$8\sqrt{3} \ m$

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(B) Let the height of the first house be $H$. Let the window be at height $y$ from the ground. The distance between the houses is $d = 6 \ m$.From the triangle formed by the bottom of the first house,the base of the second house,and the window,we have $	an(60^\circ) = \frac{y}{6}$.Thus,$y = 6 	an(60^\circ) = 6\sqrt{3} \ m$.Let the window divide the house of height $H$ into two parts $h_1$ and $h_2$ such that $H = h_1 + h_2$. The window is at height $y$ from the ground,so $h_2 = y = 6\sqrt{3} \ m$.The angle subtended by the house at the window is $90^\circ$. Let the angle of elevation of the top of the house from the window be $\alpha$. Then the angle of depression of the bottom of the house from the window is $90^\circ - \alpha$.In the upper triangle,$	an(\alpha) = \frac{h_1}{6}$. In the lower triangle,$	an(90^\circ - \alpha) = \cot(\alpha) = \frac{y}{6} = \frac{6\sqrt{3}}{6} = \sqrt{3}$.Since $\cot(\alpha) = \sqrt{3}$,we have $	an(\alpha) = \frac{1}{\sqrt{3}}$.Therefore,$h_1 = 6 	an(\alpha) = 6 	imes \frac{1}{\sqrt{3}} = 2\sqrt{3} \ m$.The total height $H = h_1 + h_2 = 2\sqrt{3} + 6\sqrt{3} = 8\sqrt{3} \ m$.

$A$ house subtends a right angle at the window of an opposite house and the angle of elevation of the window from the bottom of the first house is $60^\circ$. If the distance between the two houses is $6 \ m$,then the height of the first house is:

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