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JEE Mains · Maths · STD 11 - Trigonometrical equations

A horizontal park is in the shape of a triangle \(OAB\) with \(AB =16\). \(A\) vertical lamp post \(OP\) is erected at the point \(O\) such that \(\angle PAO =\angle PBO =15^{\circ}\) and \(\angle PCO =45^{\circ}\), where \(C\) is the midpoint of \(AB\). Then \(( OP )^{2}\) is equal to.

  1. A \(\frac{32}{\sqrt{3}}(\sqrt{3}-1)\)
  2. B \(\frac{32}{\sqrt{3}}(2-\sqrt{3})\)
  3. C \(\frac{16}{\sqrt{3}}(\sqrt{3}-1)\)
  4. D \(\frac{16}{\sqrt{3}}(2-\sqrt{3})\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{32}{\sqrt{3}}(2-\sqrt{3})\)

Step-by-step Solution

Detailed explanation

\(\frac{ OP }{ OA }=\tan 15^{\circ}\) \(OA = OP \cot 15^{\circ}\) \(\frac{ OP }{ OC }=\tan 45^{\circ} \Rightarrow OP = OC\) \(Now , OP =\sqrt{ OA ^{2}-8^{2}}\) \(OP ^{2}=( OP )^{2} \cot ^{2} 15^{\circ}-64\) \(OP ^{2}=\frac{32}{\sqrt{3}}(2-\sqrt{3})\)
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