The angle of elevation of the top P of a vert | vedclass

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(A) In $	riangle PQA$,$\angle PAQ = 45^{\circ}$ and $PQ = 10$,so $QA = \frac{PQ}{	an 45^{\circ}} = 10$.In $	riangle BRA$,$RA = d \cos 30^{\circ} =

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$(10(\sqrt{3}-1), 25)$

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(A) In $	riangle PQA$,$\angle PAQ = 45^{\circ}$ and $PQ = 10$,so $QA = \frac{PQ}{	an 45^{\circ}} = 10$.In $	riangle BRA$,$RA = d \cos 30^{\circ} = \frac{\sqrt{3}d}{2}$ and $BR = d \sin 30^{\circ} = \frac{d}{2}$.Since $R$ is on $AQ$,$QR = QA - RA = 10 - \frac{\sqrt{3}d}{2}$.In $	riangle PRB$,the angle of elevation of $P$ from $B$ is $60^{\circ}$,so $	an 60^{\circ} = \frac{PQ - BR}{QR} = \frac{10 - d/2}{10 - \sqrt{3}d/2}$.$\sqrt{3} = \frac{20 - d}{20 - \sqrt{3}d} \implies 20\sqrt{3} - 3d = 20 - d \implies 2d = 20(\sqrt{3} - 1) \implies d = 10(\sqrt{3} - 1)$.The area of trapezium $PQRB$ is $\alpha = \frac{1}{2}(PQ + BR) \cdot QR = \frac{1}{2}(10 + d/2)(10 - \sqrt{3}d/2)$.Substituting $d = 10(\sqrt{3} - 1)$,we get $BR = 5(\sqrt{3} - 1)$ and $QR = 10 - 5\sqrt{3}(\sqrt{3} - 1) = 10 - 15 + 5\sqrt{3} = 5\sqrt{3} - 5$.$\alpha = \frac{1}{2}(10 + 5\sqrt{3} - 5)(5\sqrt{3} - 5) = \frac{1}{2}(5\sqrt{3} + 5)(5\sqrt{3} - 5) = \frac{1}{2}(75 - 25) = 25$.Thus,$(d, \alpha) = (10(\sqrt{3} - 1), 25)$.

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