a) $$100\,m$$
b) $$100\sqrt 3 \,m$$
c) $$100\left( {\sqrt 3 – 1} \right)\,m$$
d) $$\frac{{100}}{{\sqrt 3 }}\,m$$

Answer: c
Explanation: Let AB be tower and AB = 100 m and angles of elevation of A at C and D are 30° and 45° respectively and CD = x
Let BD = y

$$\eqalign{ & {\text{Now in right }}\Delta ADB, \cr & \tan \theta = \frac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \frac{{AB}}{{DB}} \cr & \tan {45^ \circ } = \frac{{100}}{y} \cr & \Rightarrow 1 = \frac{{100}}{y} \cr & \Rightarrow y = 100 \cr & {\text{Similarly in right }}\Delta ACB, \cr & \tan {30^ \circ } = \frac{{AB}}{{CB}} \cr & \frac{1}{{\sqrt 3 }} = \frac{{100}}{{y + x}} \cr & \frac{1}{{\sqrt 3 }} = \frac{{100}}{{100 + x}} \cr & 100 + x = 100\sqrt 3 \cr & x = 100\sqrt 3 – 100 \cr & x = 100\left( {\sqrt 3 – 1} \right)\,m \cr} $$