a) 63.5 m
b) 76.9 m
c) 86.7 m
d) 90 m

Answer: a
Explanation:

Let AB be the tower and C and D be the objects.
Then, AB = 150 m, ∠ACB = 45° and ∠ADB = 60°
$$\eqalign{ & \frac{{AB}}{{AD}} = \tan {60^ \circ } = \sqrt 3 \cr & \Rightarrow AD = \frac{{AB}}{{\sqrt 3 }} = \frac{{150}}{{\sqrt 3 }} \cr & \frac{{AB}}{{AC}} = \tan {45^ \circ } = 1 \cr & \Rightarrow AC = AB = 150{\text{ m}} \cr & CD = \left( {AC – AD} \right) \cr & = \left( {150 – \frac{{150}}{{\sqrt 3 }}} \right){\text{m}} \cr & = \left[ {\frac{{150\left( {\sqrt 3 – 1} \right)}}{{\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }}} \right]{\text{m}} \cr & = 50\left( {3 – \sqrt 3 } \right){\text{m}} \cr & = \left( {50 \times 1.27} \right){\text{m}} \cr & = 63.5\,{\text{m}} \cr} $$