question_answer

Two men are on the opposite side of a tower. They measure the angles of elevation of the top of the tower\[{{45}^{o}}\]and\[{{30}^{o}}\]respectively. If the height of the tower is 40 m, find the distance between the men [Karnataka CET 1998]

A) 40 m

B) \[40\sqrt{3}\,m\]

C) 68.280 m

D) 109.28 m

Correct Answer: D

Solution :

From \[\Delta \text{ }{{O}_{1}}AB,\] \[\tan {{45}^{o}}=\frac{40}{x}\Rightarrow x=40m\] From \[\Delta A{{O}_{2}}B,\] \[\cot {{30}^{o}}=\frac{y}{40}\] \[\Rightarrow \] \[y=40\,\] cot \[{{30}^{o}}=40\sqrt{3}\] Distance between the men = \[40+40\sqrt{3}=109.28\] m.