Question

Let AB and PQ 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 \(θ\) be the angles of elevation from C to P and A, respectively. If the height of pole PQ is twice the height of pole AB, then \(tan^2θ\) is equal to

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

Answer 1

The Correct Option is C

\(\frac {l}{80} = tan\ θ\)          ..... (i)

\(\frac {2l}{80} = tan \frac \pi8 \)   ...... (ii)

From (i) and (ii) 

\(\frac 12 = \frac {tan\ θ}{tan \frac {\pi}{8}}\)

\(⇒ tan^2θ = \frac 14 tan^2\frac \pi 8\)

\(⇒ tan^2θ = \frac {\sqrt 2 - 1}{4(\sqrt 2 + 1)}\)

\(⇒ tan^2θ =\frac {3-2\sqrt 2}{2}\)

So, the correct option is (C): \(\frac {3-2\sqrt 2}{2}\)