Answer: Option C
Let CD is the tower and A is a point such that the angle of elevation of C is 30°
B is and their point h m high of A and angle of depression of D is 60°
$$\eqalign{
& {\text{The}}\,AB = h\,m \cr
& {\text{Let}}\,CD = H\,m\,\,\,{\text{and }}\,AD\, = x \cr
& {\text{Now}}\,{\text{in}}\,{\text{right}}\,\Delta ABD, \cr
& \tan \theta = \frac{{AB}}{{AD}} \cr
& \Rightarrow \tan {60^ \circ } = \frac{h}{x} \cr
& \Rightarrow \sqrt 3 = \frac{h}{x} \cr
& \Rightarrow x = \frac{h}{{\sqrt 3 }}\,.......({\text{i}}) \cr
& {\text{Similarly in right }}\Delta ACB, \cr
& \tan {30^ \circ } = \frac{{CD}}{{AD}} \cr
& \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{H}{x} \cr
& \Rightarrow x = \sqrt 3 \,H\,.......\left( {{\text{ii}}} \right) \cr
& {\text{From}}\,\left( {\text{i}} \right)\,{\text{and}}\,\left( {{\text{ii}}} \right) \cr
& \sqrt 3 \,H = \frac{h}{{\sqrt 3 }} \cr
& H = \frac{h}{{\sqrt 3 \times \sqrt 3 }} = \frac{h}{3} \cr
& \therefore {\text{Height of tower}} = \frac{h}{3} \cr} $$