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

Answer: b
Explanation: Let AB is a wall and AC is the ladder 15 m long which makes an angle of 60° with the ground

In ∆ABC, ∠B = 90°
Let height of wall AB = h
Then
$$\sin \theta = \frac{{AB}}{{AC}} \Rightarrow \sin {60^ \circ } = \frac{h}{{15}}$$
$$\eqalign{ & \frac{{\sqrt 3 }}{2} = \frac{h}{{15}} \cr & h = \frac{{15\sqrt 3 }}{2}\,m \cr} $$
Height of the wall $$ = \frac{{15\sqrt 3 }}{2}\,m $$