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A student is standing with a banner at the top of a 100 m high college building. From a point on the ground, the angle of elevation of the top of the student is 60$$^{\circ}$$ and from the same point, the angle of elevation of the top of the tower is 45$$^{\circ}$$. Find the height of the student?

Let the base of building be 'x' m and height of student be 'k' m
tan 45 =Β $$\ \frac{\ 100}{x}$$
x=100m
tan 60 =Β $$\frac{100+k}{100}$$
k=100($$\sqrt{3}$$ - 1) = 73.2m
Option D is the answer.
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