The angles of depression of two ships from the top of a light house are 45° and 30° towards east. If the ships are 100 m apart, the height of the light house is

Answer: Option C

Let AB be the light house C and D are two ships whose angles of depression on A are 30° and 45° respectively

∠ACB = ∠XAC = 30° , ∠ADB = ∠YAD = 45° and
CD = 100m
Let AB =h and CB = x then BC = (100 - x)m
$$\begin{array}{rl}& \text{Now in }\Delta ACB,\\ & \tan \theta =\frac{AB}{CB}\\ & \tan 30^{\circ }=\frac{h}{x}\\ & \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x}\\ & \Rightarrow x=\sqrt{3}h......\text{(i)}\\ & \text{Similarly in }\Delta ADB\\ & \tan 45^{\circ }=\frac{AB}{BD}\\ & \Rightarrow 1=\frac{h}{100-x}\\ & \Rightarrow x=100-h......\text{(ii)}\\ & \text{From (i) and (ii)}\\ & \sqrt{3}h=100-h\\ & \Rightarrow (\sqrt{3}-1)h=100\\ & h=\frac{100}{\sqrt{3}+1}\\ & =\frac{100\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\\ & =\frac{100\left(\sqrt{3}-1\right)}{3-1}\\ & =\frac{100\left(\sqrt{3}-1\right)}{2}\\ & =50\left(\sqrt{3}-1\right)\end{array}$$
∴ height of light house $$=50\left(\sqrt{3}-1\right)m$$