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, \[A{{Q}^{2}}+B{{Q}^{2}}={{100}^{2}}\] \[\Rightarrow \,{{h}^{2}}+{{h}^{2}}{{\cot }^{2}}60{}^\circ ={{100}^{2}}\] \[\Rightarrow \,\,{{h}^{2}}\left[ 1+\frac{1}{3} \right]={{100}^{2}}\] \[\Rightarrow \,\,{{h}^{2}}=\frac{3\times {{100}^{2}}}{4}\Rightarrow h=50\sqrt{3}\]You need to login to perform this action.
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