From an aeroplane vertically over a straight | vedclass

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Question

(A) Let $H$ be the height of the aeroplane above the road in miles.Let $P$ be the position of the aeroplane and $Q$ be the point on the road directly

Vedclass · Accepted Answer

$\frac{\sqrt{3}}{2}$

Vedclass · Answer

(A) Let $H$ be the height of the aeroplane above the road in miles.Let $P$ be the position of the aeroplane and $Q$ be the point on the road directly below it.Let $R$ and $S$ be the positions of the two consecutive milestones on the road.The distance between two consecutive milestones is $RS = 1 	ext{ mile}$.In $\Delta PQR$,$	an 60^{\circ} = \frac{H}{QR} \Rightarrow QR = \frac{H}{\sqrt{3}}$.In $\Delta PQS$,$	an 30^{\circ} = \frac{H}{QS} \Rightarrow QS = H\sqrt{3}$.Since $QS = QR + RS$,we have $H\sqrt{3} = \frac{H}{\sqrt{3}} + 1$.Multiplying by $\sqrt{3}$,we get $3H = H + \sqrt{3}$.$2H = \sqrt{3} \Rightarrow H = \frac{\sqrt{3}}{2} 	ext{ miles}$.

From an aeroplane vertically over a straight horizontal road,the angles of depression of two consecutive milestones on the opposite sides of the aeroplane are observed to be $30^{\circ}$ and $60^{\circ}$. Find the height in miles of the aeroplane above the road.

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