question_answer

Two men are on opposite sides of a tower. The measure the angles of elevation of the top of di tower as \[30{}^\circ \]and \[45{}^\circ ,\]respectively. If the height of a tower is 50 m, find the distance between the two men. (Take\[\sqrt{3}=1.73\])

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) \[=(50\sqrt{3}+50)\,\,m=50(\sqrt{3}+1)\] \[=50(1.73+1)\,\,m=(50\times 2.73)\,\,m=136.5\,\,m\]