A man on the top of a vertical observation to | vedclass

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(C) Let $PQ$ be the height of the tower $H$. Let the car be at point $S$ initially and at point $R$ after $12$ minutes. The angles of depression are $

Vedclass · Accepted Answer

$16 \min 23 \sec$

Vedclass · Answer

(C) Let $PQ$ be the height of the tower $H$. Let the car be at point $S$ initially and at point $R$ after $12$ minutes. The angles of depression are $30^{\circ}$ and $45^{\circ}$ respectively,which are equal to the angles of elevation $\angle PSQ = 30^{\circ}$ and $\angle PRQ = 45^{\circ}$.In $\Delta PQR$,$	an 45^{\circ} = \frac{PQ}{RQ} = 1 \implies PQ = RQ = H$.In $\Delta PQS$,$	an 30^{\circ} = \frac{PQ}{SQ} = \frac{1}{\sqrt{3}} \implies SQ = H\sqrt{3}$.The distance $SR = SQ - RQ = H\sqrt{3} - H = H(\sqrt{3} - 1)$.The car covers distance $SR$ in $12$ minutes. Let the time taken to cover distance $RQ$ be $t$ minutes.Since speed is uniform,$\frac{SR}{12} = \frac{RQ}{t}$.$\frac{H(\sqrt{3} - 1)}{12} = \frac{H}{t} \implies t = \frac{12}{\sqrt{3} - 1} = \frac{12(\sqrt{3} + 1)}{3 - 1} = 6(\sqrt{3} + 1) \approx 6(1.732 + 1) = 6(2.732) = 16.392 	ext{ minutes}$.$0.392 	ext{ minutes} = 0.392 	imes 60 \approx 23.5 	ext{ seconds}$.Thus,the time is $16 	ext{ minutes } 23 	ext{ seconds}$.

$A$ man on the top of a vertical observation tower observes a car moving at a uniform speed coming directly towards it. If it takes $12$ minutes for the angle of depression to change from $30^{\circ}$ to $45^{\circ},$ how soon after this will the car reach the observation tower?

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