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Question

(A) Let $PQ$ be the tower and $R$ and $S$ be the two positions of the boat. Given $RQ = 60 \ m$,$\angle PSQ = 30^{\circ}$,and $\angle PRQ = 45^{\circ}

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(A) Let $PQ$ be the tower and $R$ and $S$ be the two positions of the boat. Given $RQ = 60 \ m$,$\angle PSQ = 30^{\circ}$,and $\angle PRQ = 45^{\circ}$.From $\Delta PQR$,$	an 45^{\circ} = \frac{PQ}{RQ} = 1$.Thus,$PQ = RQ = 60 \ m$.From $\Delta PQS$,$	an 30^{\circ} = \frac{PQ}{QS} = \frac{60}{RQ + RS} = \frac{60}{60 + RS} = \frac{1}{\sqrt{3}}$.This implies $60 + RS = 60\sqrt{3}$.So,$RS = 60(\sqrt{3} - 1) \ m$.Using $\sqrt{3} \approx 1.732$,$RS = 60(1.732 - 1) = 60 	imes 0.732 = 43.92 \ m$.Speed of the boat = $\frac{	ext{Distance}}{	ext{Time}} = \frac{43.92 \ m}{5 \ s} = 8.784 \ m/s$.To convert to $km/hr$,multiply by $\frac{18}{5}$:Speed = $8.784 	imes \frac{18}{5} = 31.62 \ km/hr$.Rounding to the nearest integer,the speed is approximately $32 \ km/hr$.

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