question_answer

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 \[\text{45 }\!\!{}^\circ\!\!\text{ }\] , \[\text{30 }\!\!{}^\circ\!\!\text{ }\] and \[\text{30 }\!\!{}^\circ\!\!\text{ }\], then the height of the tower (in m) is:                    [JEE Main Online 08-04-2018]

A)  \[\text{100}\sqrt{3}\]            

B)  \[50\sqrt{2}\]

C)  100                            

D)  50

Correct Answer: C

Solution :

Let AB=h, \[\text{QR=2a}\] \[\text{In }\Delta \text{ABQ, tan30}{}^\circ \text{=}\frac{AB}{QB}=\frac{h}{a}\Rightarrow a=h\sqrt{3}\] \[\text{In }\Delta \text{PBA, tan45}{}^\circ \text{=}\frac{AB}{PB}=\frac{h}{\sqrt{{{200}^{2}}-{{a}^{2}}}}\]             \[\Rightarrow 1=\frac{h}{\sqrt{{{200}^{2}}-3{{h}^{2}}}}\] \[\Rightarrow 4{{h}^{2}}={{200}^{2}}\] \[\Rightarrow h=100\]