a) $$\frac{{50}}{{\sqrt 3 }}\left( {\sqrt 3 – 1} \right)$$
b) $$10\sqrt 3 \,{\text{m}}$$
c) $$50\sqrt 3 \,{\text{m}}$$
d) None of these

Answer: a
Explanation:

$$\eqalign{ & {\text{in}}\,\Delta PNQ,\,\tan {45^ \circ } = \frac{{PQ}}{{NQ}} \cr & PQ = NQ = 50{\text{m}} \cr & {\text{in}}\,\Delta MNQ,\,\tan {30^ \circ } = \frac{{MQ}}{{NQ}} \cr & \frac{1}{{\sqrt 3 }} = \frac{{MQ}}{{50}} \cr & MQ = \frac{{50}}{{\sqrt 3 }} \cr & h = PM \cr & \,\,\,\,\,\,\,\,\,\, = PQ – MQ \cr & \,\,\,\,\,\,\,\,\,\, = 50 – \frac{{50}}{{\sqrt 3 }} \cr & \,\,\,\,\,\,\,\,\,\, = \frac{{50}}{{\sqrt 3 }}\left( {\sqrt 3 – 1} \right) \cr} $$
Poster height = $$\frac{{50}}{{\sqrt 3 }}\left( {\sqrt 3 – 1} \right)$$