An observer on the top of a cliff,200 , m abo | vedclass

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(D) Let the height of the cliff be $PS = 200 \, m$. Let the two ships be at points $Q$ and $R$ such that $S, Q, R$ are collinear.In $\Delta PQS$,the a

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$146.4$

Vedclass · Answer

(D) Let the height of the cliff be $PS = 200 \, m$. Let the two ships be at points $Q$ and $R$ such that $S, Q, R$ are collinear.In $\Delta PQS$,the angle of elevation is $45^{\circ}$.$	an 45^{\circ} = \frac{PS}{SQ} \implies 1 = \frac{200}{SQ} \implies SQ = 200 \, m$.In $\Delta PRS$,the angle of elevation is $30^{\circ}$.$	an 30^{\circ} = \frac{PS}{SR} \implies \frac{1}{\sqrt{3}} = \frac{200}{SQ + QR}$.$SQ + QR = 200\sqrt{3}$.$200 + QR = 200(1.732) = 346.4$.$QR = 346.4 - 200 = 146.4 \, m$.Thus,the distance between the two ships is $146.4 \, m$.

An observer on the top of a cliff,$200 \, m$ above the sea level,observes the angle of depression of two ships at anchor to be $45^{\circ}$ and $30^{\circ}$ respectively. Find the distance (in $m$) between the ships if the line joining them stretches to the base of the cliff.

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