The angle of depression of a car parked on th | vedclass

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Question

(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 i

Vedclass · Accepted Answer

$20$

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}) = H/x$. This gives $H = x \sqrt{3}$. From the window $14.6 \, m$ below the top,the height is $(H - 14.6)$. The angle of depression is $45^{\circ}$,so $	an(45^{\circ}) = (H - 14.6)/x$. Since $	an(45^{\circ}) = 1$,we have $x = H - 14.6$,which means $H = x + 14.6$. Equating the two expressions for $H$: $x \sqrt{3} = x + 14.6$. $x(\sqrt{3} - 1) = 14.6$. $x = 14.6 / (1.732 - 1) = 14.6 / 0.732 \approx 19.945 \, m$. Rounding to the nearest integer,the distance is $20 \, m$.

The angle of depression of a car parked on the road as observed from the top of a building is $60^{\circ}$. The angle of depression of that car from a window $14.6 \, m$ below the top is $45^{\circ}$. Find the distance between the car and the building (in $m$).

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