a) $$\sqrt {mn\sqrt 3 } \,{\text{units}}$$
b) $$\sqrt {mn\root 4 \of 3 } \,{\text{units}}$$
c) $$\sqrt {3mn} \,{\text{units}}$$
d) $$\sqrt {mn} \,{\text{units}}$$

Answer: a
Explanation:

$$\eqalign{ & {\text{Height}}\,{\text{of}}\,{\text{pole}} = PQ \cr & \tan {60^ \circ } = \sqrt 3 = \frac{{PQ}}{m} \cr & \tan {45^ \circ } = 1 = \frac{{PQ}}{n} \cr & {\text{Multiply}}\,{\text{both}}\,{\text{equations}} \cr & \sqrt 3 \times 1 = \frac{{PQ}}{m} \times \frac{{PQ}}{n} \cr & PQ = \sqrt {mn\sqrt 3 } \,{\text{units}} \cr} $$