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Question

(A) Let $AQ = d$. In $	riangle AQR$,$	an 60^{\circ} = \frac{QR}{AQ} = \frac{15}{d}$.So,$d = \frac{15}{	an 60^{\circ}} = \frac{15}{\sqrt{3}} = 5\sqr

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$5(2 \sqrt{3} + 3) \, m$

Vedclass · Answer

(A) Let $AQ = d$. In $	riangle AQR$,$	an 60^{\circ} = \frac{QR}{AQ} = \frac{15}{d}$.So,$d = \frac{15}{	an 60^{\circ}} = \frac{15}{\sqrt{3}} = 5\sqrt{3} \, m$.In $	riangle AQP$,the angle of elevation of $P$ is $60^{\circ} + 15^{\circ} = 75^{\circ}$.Thus,$	an 75^{\circ} = \frac{PQ}{AQ} = \frac{15 + x}{5\sqrt{3}}$,where $x = PR$.Since $	an 75^{\circ} = 	an(45^{\circ} + 30^{\circ}) = \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} = \frac{(\sqrt{3} + 1)^2}{3 - 1} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$.Therefore,$15 + x = 5\sqrt{3}(2 + \sqrt{3}) = 10\sqrt{3} + 15$.This gives $x = 10\sqrt{3} \, m$.The total height of the tower $PQ = QR + PR = 15 + 10\sqrt{3} = 5(3 + 2\sqrt{3}) \, m$.

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