JEE Mains · Maths · STD 11 - Trigonometrical equations
A tower \(P Q\) stands on a horizontal ground with base \(Q\) on the ground. The point \(R\) divides the tower in two parts such that \(QR =15\,m\). If from a point \(A\) on the ground the angle of elevation of \(R\) is \(60^{\circ}\) and the part \(PR\) of the tower subtends an angle of \(15^{\circ}\) at \(A\), then the height of the tower is.
- A \(5(2 \sqrt{3}+3)\,m\)
- B \(5(\sqrt{3}+3)\,m\)
- C \(10(\sqrt{3}+1)\,m\)
- D \(10(2 \sqrt{3}+1)\,m\)
Answer & Solution
Correct Answer
(A) \(5(2 \sqrt{3}+3)\,m\)
Step-by-step Solution
Detailed explanation
\(\frac{15}{ AQ }=\tan 60^{\circ}\) \(\frac{15+ x }{ AQ }=\tan 75^{\circ}\) \(\frac{(1)}{(2)} \Rightarrow x =10 \sqrt{3}\) So, \(PQ =5(2 \sqrt{3}+3)\,m\)
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