# CAT Practice : Trigonometry

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Max and Min value of the given expression

## Heights and Distances

Q.12: 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° and from the same point, the angle of elevation of the top of the tower is 45°. Find the height of the student.
1. 35 m
2. 73.2 m
3. 50 m
4. 75 m

Choice B. 73.2 m

## Detailed Solution

Let BC be the height of the tower and DC be the height of the student.
In rt. ∆ ABC
AB = BC cot 45°
AB = 100 m ……………(1)
In rt. ∆ ABD
AB = BD cot 60°
AB = (BC + CD) cot 60°
AB = (10 + CD) * $\frac{1}{√3}$
Equating (i) and (ii)
(10 + CD) * $\frac{1}{√3}$ = 100
(10 + CD)= 100√3
CD = 100√3 – 100
= 10(1.732 – 1) = 100 x 0.732 = 73.2 m

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