a) $$\frac{{50}}{{\sqrt 3 + 1}}m$$
b) $$\frac{{50}}{{\sqrt 3 – 1}}m$$
c) $$50\left( {\sqrt 3 – 1} \right)m$$
d) $$50\left( {\sqrt 3 + 1} \right)m$$

Answer: c
Explanation: 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
$$\eqalign{ & {\text{Now in }}\,\Delta ACB, \cr & \tan \theta = \frac{{AB}}{{CB}} \cr & \tan {30^ \circ } = \frac{h}{x} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{h}{x} \cr & \Rightarrow x = \sqrt 3 h ……{\text{(i)}} \cr & {\text{Similarly in }}\,\Delta ADB \cr & \tan {45^ \circ } = \frac{{AB}}{{BD}} \cr & 1 = \frac{h}{{100 – x}} \cr & x = 100 – h \,……{\text{(ii)}} \cr & {\text{From (i) and (ii)}} \cr & \sqrt 3 h = 100 – h \cr & \Rightarrow (\sqrt 3 – 1)h = 100 \cr & h = \frac{{100}}{{\sqrt 3 + 1}} \cr & \,\,\,\,\,\, = \frac{{100\left( {\sqrt 3 – 1} \right)}}{{\left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 – 1} \right)}} \cr & \,\,\,\,\,\, = \frac{{100\left( {\sqrt 3 – 1} \right)}}{{3 – 1}} \cr & \,\,\,\,\,\, = \frac{{100\left( {\sqrt 3 – 1} \right)}}{2} \cr & \,\,\,\,\,\, = 50\left( {\sqrt 3 – 1} \right) \cr} $$
height of light house $$ = 50\left( {\sqrt 3 – 1} \right)m$$