JEE Mains · Maths · STD 11 - Trigonometrical equations
\(PQR\) triangular park with \(PQ = PR = 200\ m.A\) TV tower stands at the mid-point of \(QR\). If the angles of elevation of the top of the tower at \(P, Q\) and \(R\) are respectively \(45^o , 30^o \) and \(30^o \), then the height of the tower \((in \,m)\) is:
- A \(50\)
- B \(100\sqrt 3 \)
- C \(50\sqrt 2 \;\)
- D \(100\)
Answer & Solution
Correct Answer
(D) \(100\)
Step-by-step Solution
Detailed explanation
Let hight of tower \(MN=h\) In \(\Delta QMN\) we have \(\tan 30^o=\frac {MN}{QM}\) \(\therefore QM\,=\,\sqrt 3 h=MR\) ....... \((1)\) Now in \(\Delta MNP\) \(MN=PM\) ........ \((2)\) In \(\Delta PMQ\) we have : \(MP\,=\,\sqrt {(200)^2-(\sqrt 3h)^2}\) \(\therefore \) From \((2)\)…
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