JEE Mains · Maths · STD 11 - Trigonometrical equations
The angle of elevation of the top \(P\) of a vertical tower \(PQ\) of height \(10\) from a point \(A\) on the horizontal ground is \(45^{\circ}\). Let \(R\) be a point on \(AQ\) and from a point \(B\), vertically above \(R\), the angle of elevation of \(P\) is \(60^{\circ}\). If \(\angle BAQ =30^{\circ}, AB = d\) and the area of the trapezium \(PQRB\) is \(\alpha\), then the ordered pair \(( d , \alpha)\) is.
- A \((10(\sqrt{3}-1), 25)\)
- B \(\left(10(\sqrt{3}-1), \frac{25}{2}\right)\)
- C \((10(\sqrt{3}+1), 25)\)
- D \(\left(10(\sqrt{3}+1), \frac{25}{2}\right)\)
Answer & Solution
Correct Answer
(A) \((10(\sqrt{3}-1), 25)\)
Step-by-step Solution
Detailed explanation
\(QA =10 \quad RA = d \cos 30^{\circ}=\frac{\sqrt{3} d }{2}\) \(QR =10-\frac{\sqrt{3} d }{2}\) \(BR = d \sin 30^{\circ}=\frac{ d }{2}\) \(\tan 60^{\circ}=\frac{ PQ - BR }{ QR }=\frac{10-\frac{ d }{2}}{10-\frac{\sqrt{3} d }{2}}\) \(\sqrt{3}=\frac{20- d }{20-\sqrt{3} d }\)…
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