A house subtends a right angle at the window | vedclass

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Question

(A) Let $AB$ be the distance between the two houses,$AB = 6 \ m$. Let $D$ be the position of the window on the second house,and $BC$ be the height of

Vedclass · Accepted Answer

$8\sqrt{3} \ m$

Vedclass · Answer

(A) Let $AB$ be the distance between the two houses,$AB = 6 \ m$. Let $D$ be the position of the window on the second house,and $BC$ be the height of the first house,$BC = h$.Given that the angle of elevation of the window $D$ from the bottom $B$ of the first house is $30^{\circ}$,so $\angle DBA = 30^{\circ}$.In $\Delta ABD$,$	an 30^{\circ} = \frac{AD}{AB} \implies \frac{1}{\sqrt{3}} = \frac{AD}{6} \implies AD = \frac{6}{\sqrt{3}} = 2\sqrt{3} \ m$.Also,$\cos 30^{\circ} = \frac{AB}{BD} \implies \frac{\sqrt{3}}{2} = \frac{6}{BD} \implies BD = \frac{12}{\sqrt{3}} = 4\sqrt{3} \ m$.The house subtends a right angle at the window,so $\angle BDC = 90^{\circ}$.In $\Delta BDC$,$\angle DBC = 60^{\circ}$ (since $\angle ABC = 90^{\circ}$ and $\angle ABD = 30^{\circ}$,$\angle DBC = 90^{\circ} - 30^{\circ} = 60^{\circ}$).Then $\cos 60^{\circ} = \frac{BD}{BC} \implies \frac{1}{2} = \frac{4\sqrt{3}}{h}$.Therefore,$h = 8\sqrt{3} \ m$.

$A$ house subtends a right angle at the window of an opposite house. The angle of elevation of the window from the bottom of the first house is $30^{\circ}$. If the distance between the two houses is $6 \ m$,then the height of the first house is:

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