On the same side of tower, two objects are located. Observed from the top of the tower, their angles of depression are 45° and 60°. If the height of the tower is 150 m, the distance between the objects is-

Solution(By Examveda Team)

Let AB be the tower and C and D be the objects.
Then, AB = 150 m, ∠ACB = 45° and ∠ADB = 60°
$$\eqalign{ & \frac{{AB}}{{AD}} = \tan {60^ \circ } = \sqrt 3 \cr & \Rightarrow AD = \frac{{AB}}{{\sqrt 3 }} = \frac{{150}}{{\sqrt 3 }} \cr & \frac{{AB}}{{AC}} = \tan {45^ \circ } = 1 \cr & \Rightarrow AC = AB = 150{\text{ m}} \cr & \therefore CD = \left( {AC - AD} \right) \cr & = \left( {150 - \frac{{150}}{{\sqrt 3 }}} \right){\text{m}} \cr & = \left[ {\frac{{150\left( {\sqrt 3 - 1} \right)}}{{\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }}} \right]{\text{m}} \cr & = 50\left( {3 - \sqrt 3 } \right){\text{m}} \cr & = \left( {50 \times 1.27} \right){\text{m}} \cr & = 63.5\,{\text{m}} \cr} $$