a) $$30\sqrt 3 \,m$$
b) $$15\,m$$
c) $$\frac{{30}}{{\sqrt 3 }}\,m$$
d) $$15\sqrt 2 \,m$$

Answer: a
Explanation: Let AB be tower and a point P distance of 30 m from its foot of the tower which form an angle of elevation pf the sun of 60°

$$\eqalign{ & {\text{Let height of tower }}AB = h \cr & {\text{Then in right }}\Delta APB, \cr & \tan \theta = \frac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \frac{{AB}}{{PB}} \cr & \Rightarrow \tan {60^ \circ } = \frac{h}{{30}} \cr & \sqrt 3 = \frac{h}{{30}} \cr & h = 30\sqrt 3 \cr} $$
Height of the tower $$ = 30\sqrt 3 \,m$$