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(B) Let the height of the building be $H$ and the distance of the car from the building be $x$.From the top of the building,the angle of depression is

Vedclass · Accepted Answer

$13.70$

Vedclass · Answer

(B) Let the height of the building be $H$ and the distance of the car from the building be $x$.From the top of the building,the angle of depression is $60^{\circ}$,so $	an(60^{\circ}) = \frac{H}{x} \implies \sqrt{3} = \frac{H}{x} \implies H = x\sqrt{3}$.From a window $10 \, m$ below the top,the height is $(H - 10)$ and the angle of depression is $45^{\circ}$,so $	an(45^{\circ}) = \frac{H - 10}{x} \implies 1 = \frac{H - 10}{x} \implies H - 10 = x$.Substitute $H = x\sqrt{3}$ into the second equation: $x\sqrt{3} - 10 = x$.$x(\sqrt{3} - 1) = 10$.$x = \frac{10}{\sqrt{3} - 1} = \frac{10}{1.732 - 1} = \frac{10}{0.732} \approx 13.66 \, m$.

Watching from the top of a building and from a window $10 \, m$ below the top,the angles of depression of a car on the ground are found to be $60^{\circ}$ and $45^{\circ}$ respectively. Find the distance of the car from the building. (in $m$)

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