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-

Answer: Option A

Solution

Answer: Option A

Solution:

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