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Question

(A) Let $AB = 30 \ m$. Let $BQ = x$. In $	riangle ABQ$,$	an 60^{\circ} = \frac{AB}{BQ} = \frac{30}{x}$.So,$x = \frac{30}{\sqrt{3}} = 10\sqrt{3} \ m$

Vedclass · Accepted Answer

$600(\sqrt{3}-1)$

Vedclass · Answer

(A) Let $AB = 30 \ m$. Let $BQ = x$. In $	riangle ABQ$,$	an 60^{\circ} = \frac{AB}{BQ} = \frac{30}{x}$.So,$x = \frac{30}{\sqrt{3}} = 10\sqrt{3} \ m$. Thus,$BQ = 10\sqrt{3} \ m$.Since $BCPQ$ is a rectangle,$CP = BQ = 10\sqrt{3} \ m$ and $PQ = BC$.In $	riangle ACP$,the angle of depression of $P$ is $15^{\circ}$,so $\angle ACP = 90^{\circ}$ and $\angle CAP = 90^{\circ} - 15^{\circ} = 75^{\circ}$ is not correct,rather $\angle APC = 15^{\circ}$ (alternate interior angle).Thus,$	an 15^{\circ} = \frac{AC}{CP} = \frac{AC}{10\sqrt{3}}$.Since $	an 15^{\circ} = 2 - \sqrt{3}$,we have $AC = 10\sqrt{3}(2 - \sqrt{3}) = 20\sqrt{3} - 30$.Then $BC = AB - AC = 30 - (20\sqrt{3} - 30) = 60 - 20\sqrt{3}$.Area of rectangle $BCPQ = BQ 	imes BC = (10\sqrt{3})(60 - 20\sqrt{3}) = 600\sqrt{3} - 200(3) = 600\sqrt{3} - 600 = 600(\sqrt{3} - 1) \ m^2$.

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