question_answer

The angle of elevation of the top of vertical tower standing on a horizontal plane is observed to be \[45{}^\circ \]from a point A on the plane. Let B be the point 30 m vertically above the point A. If the angle of elevation of the top of the tower from B be \[30{}^\circ ,\]then the distance (in m) of the foot of the tower from the point A is: [JEE Main 12-4-2019 Afternoon]

A) \[15\left( 3-\sqrt{3} \right)\]                 

B) \[15\left( 3+\sqrt{3} \right)\]

C) \[15\left( 1+\sqrt{3} \right)\]               

D) \[15\left( 5-\sqrt{3} \right)\]

Correct Answer: B

Solution :

\[\tan {{45}^{o}}=1=\frac{x+30}{y}\Rightarrow x+30=y\]            (i) \[\tan {{30}^{o}}=\frac{1}{\sqrt{3}}=\frac{x}{y}\Rightarrow x=\frac{y}{\sqrt{3}}\]                  (ii) from (i) and (ii) \[y=15\left( 3+\sqrt{3} \right)\]