a) 8.65 m
b) 2 m
c) 2.5 m
d) 3.65 m

Answer: d
Explanation:

Let AB be the man and CD be the window
Given that the height of the man, AB = 180 cm, the distance between the man and the wall, BE = 5 m,
∠ DAF = 45°, ∠ CAF = 60°
From the diagram, AF = BE = 5 m
$$\eqalign{ & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta \,AFD, \cr & \tan {45^ \circ } = \frac{{DF}}{{AF}} \cr & 1 = \frac{{DF}}{5} \cr & DF = 5\,……\left( 1 \right) \cr & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta \,AFC, \cr & \tan {60^ \circ } = \frac{{CF}}{{AF}} \cr & \sqrt 3 = \frac{{CF}}{5} \cr & CF = 5\sqrt 3 \,……\,\left( 2 \right) \cr & {\text{Length}}\,{\text{of}}\,{\text{the}}\,{\text{window}} \cr & = CD = \left( {CF – DF} \right) \cr} $$
  $$ = 5\sqrt 3 – 5$$     [ Substituted the value of CF and DF from (1) and (2)]
$$\eqalign{ & = 5\left( {\sqrt 3 – 1} \right) \cr & = 5\left( {1.73 – 1} \right) \cr & = 5 \times 0.73 \cr & = 3.65\,{\text{m}} \cr} $$