JEE Mains · Maths · STD 11 - Trigonometrical equations
Let \(A B\) and \(P Q\) be two vertical poles, \(160\, m\) apart from each other. Let \(C\) be the middle point of \(B\) and \(Q\), which are feet of these two poles. Let \(\frac{\pi}{8}\) and \(\theta\) be the angles of elevation from \(C\) to \(P\) and \(A\), respectively. If the height of pole \(P Q\) is twice the height of pole \(AB\), then \(\tan ^{2} \theta\) is equal to
- A \(\frac{3-2 \sqrt{2}}{2}\)
- B \(\frac{3+\sqrt{2}}{2}\)
- C \(\frac{3-2 \sqrt{2}}{4}\)
- D \(\frac{3-\sqrt{2}}{4}\)
Answer & Solution
Correct Answer
(C) \(\frac{3-2 \sqrt{2}}{4}\)
Step-by-step Solution
Detailed explanation
Let \(BC = CQ = x\) and \(AB = h\) and \(PQ =2 h\) \(\tan \theta=\frac{h}{x},\tan \frac{\pi}{8}=\frac{2 h}{x}\) \(\frac{\tan \theta}{\tan \left(\frac{\pi}{8}\right)}=\frac{1}{2}\) \(\tan \theta=\frac{1}{2} \tan \left(\frac{\pi}{8}\right)=\frac{1}{2}(\sqrt{2}-1)\)…
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