question_answer

A tower standing at right angles to the ground subtends angles \[{{\sin }^{-1}}\frac{1}{3}\] and \[{{\sin }^{-1}}\frac{1}{\sqrt{5}}\] at two points A and B situated in a line through the foot of the tower and on the opposite sides. If \[AB=50\] units, then the height of the tower is:

A) 50

B) \[25\sqrt{2}\]

C) \[50(\sqrt{6}-2)\]

D) \[25(\sqrt{2}-1)\]

Correct Answer: D

Solution :

[d] \[{{\sin }^{-1}}\frac{1}{3}={{\cot }^{-1}}2\sqrt{2}\] and \[{{\sin }^{-1}}\frac{1}{\sqrt{5}}={{\cot }^{-1}}2\] If C is the foot of the lower and h is the height, then \[AC=h.2\sqrt{2},\,\,CB=h.2,\,\,h(2\sqrt{2}+2)=50\] \[\Rightarrow h=25(\sqrt{2}-1)\]