question_answer

ABC is a triangular park with \[AB=AC=100\text{ }m.\]A TV tower stands at the mid-point of BC. The angles of elevation of the top of the tower at A, B, C are \[45{}^\circ ,\text{ }60{}^\circ ,\text{ }60{}^\circ \]respectively The height of the tower is

A) \[50\,m\]

B)        \[50\,\sqrt{3}\,m\]

C) \[50\,\sqrt{2}\,m\]

D)        \[50\,(3-\sqrt{3})\,m\]

Correct Answer: B

Solution :

Let A be the height of tower, \[\tan 45{}^\circ =\frac{PQ}{AQ}=\frac{h}{AQ}\Rightarrow h=AQ\] Where PQ is tower and ABC is the park, with Q being mid point of the side BC and \[PQ=h\] Also, \[A{{Q}^{2}}+B{{Q}^{2}}={{100}^{2}}\] \[\Rightarrow \,{{h}^{2}}+{{h}^{2}}{{\cot }^{2}}60{}^\circ ={{100}^{2}}\] \[\Rightarrow \,\,{{h}^{2}}\left[ 1+\frac{1}{3} \right]={{100}^{2}}\] \[\Rightarrow \,\,{{h}^{2}}=\frac{3\times {{100}^{2}}}{4}\Rightarrow h=50\sqrt{3}\]