JEE Mains · Maths · STD 11 - Trigonometrical equations
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 \(P Q\) are \(15^{\circ}\) and \(60^{\circ}\) respectively. \(B\) and \(Q\) are on the same horizontal level. If \(C\) is a point on \(A B\) such that \(C B=P Q\), then the area (in \(m ^2\) ) of the quadrilateral \(BCPQ\) is equal to
- A \(600(\sqrt{3}-1)\)
- B \(300(\sqrt{3}+1)\)
- C \(200(3-\sqrt{3})\)
- D \(300(\sqrt{3}-1)\)
Answer & Solution
Correct Answer
(A) \(600(\sqrt{3}-1)\)
Step-by-step Solution
Detailed explanation
\(\tan 60^{\circ}=\sqrt{3}=\frac{30}{ BQ }\) \(BQ =10 \sqrt{3} m = CP\) \(\tan 15^{\circ}=2-\sqrt{3}=\frac{ AC }{ CP }\) \(AC =10 \sqrt{3}(2-\sqrt{3})\) \(\text { Area }=10 \sqrt{3}(60-20 \sqrt{3})=600(\sqrt{3}-1)\)
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