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From the top A of a vertical wall AB of height 30 m, the angles of depression of the top P and bottom Q of a vertical tower PQ are 15° and 60° respectively, B and Q are on the same horizontal level. If C is a point on AB such that CB = PQ, then the area (in m2 ) of the quadrilateral BCPQ is equal to :

Option: 1

200(3-\sqrt{3})


Option: 2

300(\sqrt{3}+1)


Option: 3

300(\sqrt{3}-1)


Option: 4

600(\sqrt{3}-1)


Answers (1)

\triangle \mathrm{ABQ}

\frac{\mathrm{AB}}{\mathrm{BQ}}=\tan 60^{\circ}
\mathrm{BQ}=\frac{30}{\sqrt{3}}=10 \sqrt{3}=\mathrm{y}

\& \triangle \mathrm{ACP}
\frac{\mathrm{AC}}{\mathrm{CP}}=\tan 15^{\circ} \Rightarrow \frac{(30-x)}{\mathrm{y}}=(2-\sqrt{3})
30-x=10 \sqrt{3}(2-\sqrt{3})
30-x=20 \sqrt{3}-30
x=60-20 \sqrt{3}

\text { Area }=x \cdot y=20(3-\sqrt{3}) \cdot 10 \sqrt{3}
=600(\sqrt{3}-1)

Ans. (4)

Posted by

Sumit Saini

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