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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

  1. A \(\frac{3-2 \sqrt{2}}{2}\)
  2. B \(\frac{3+\sqrt{2}}{2}\)
  3. C \(\frac{3-2 \sqrt{2}}{4}\)
  4. D \(\frac{3-\sqrt{2}}{4}\)
Verified Solution

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 \(PQ =2 h\) \(\tan \theta=\frac{h}{x},\tan \frac{\pi}{8}=\frac{2 h}{x}\) \(\frac{\tan \theta}{\tan \left(\frac{\pi}{8}\right)}=\frac{1}{2}\) \(\tan \theta=\frac{1}{2} \tan \left(\frac{\pi}{8}\right)=\frac{1}{2}(\sqrt{2}-1)\)…