A) 115.6 m
B) 112.5 m
C) 125.4 m
D) 136. 5 m
Correct Answer: D
Solution :
Let AB be the tower and let C and D be the positions two men, Than, \[\angle ACB=30{}^\circ ,\]\[\angle ADB=45{}^\circ \]and \[AB=50\,\,m.\] \[\frac{AC}{AB}=\cot 30{}^\circ =\sqrt{3}\]\[\Rightarrow \]\[\frac{AC}{50}=\sqrt{3}\] \[\Rightarrow \]\[AC=50\sqrt{3}\,\,m\] \[\frac{AD}{AB}=\cot 45{}^\circ =1\]\[\Rightarrow \]\[\frac{AD}{50}=1\]\[\Rightarrow \]\[AD=50\,\,m\] Distance between the two men = CD = s (AC + AD) \[=(50\sqrt{3}+50)\,\,m=50(\sqrt{3}+1)\] \[=50(1.73+1)\,\,m=(50\times 2.73)\,\,m=136.5\,\,m\]You need to login to perform this action.
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