JEE Main On 08 April 2018 Question 27
Question: PQR is a triangular park with PQ=PR=200 m. A T.V. 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 $ \text{45 }{}^\circ $ , $ \text{30 }{}^\circ $ and $ \text{30 }{}^\circ $ , then the height of the tower (in m) is: [JEE Main Online 08-04-2018]
Options:
A) $ 100\sqrt{3} $
B) $ 50\sqrt{2} $
C) 100
D) 50
Show Answer
Answer:
Correct Answer: C
Solution:
Let AB=h, $ \text{QR=2a} $
$ \text{In }\Delta \text{ABQ, tan30}{}^\circ \text{=}\frac{AB}{QB}=\frac{h}{a}\Rightarrow a=h\sqrt{3} $
$ \text{In }\Delta \text{PBA, tan45}{}^\circ \text{=}\frac{AB}{PB}=\frac{h}{\sqrt{200^{2}-a^{2}}} $
$ \Rightarrow 1=\frac{h}{\sqrt{200^{2}-3h^{2}}} $
$ \Rightarrow 4h^{2}=200^{2} $
$ \Rightarrow h=100 $
