The question is from CAT Geometry - Trigonometry. This is about Heights and Distances. We need to find out the height of a student who is standing at a top of building. Trigonometry is an important topic for CAT Preparation for the CAT Exam. Trigonometric ideas can be completely opposite, as in, some questions test lot of common sense with direction, heights and distances, while others could test you on identities. CAT exam could test one on both these fronts. Make sure you master both in CAT - Trigonometry.

Question 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.

- 35 m
- 73.2 m
- 50 m
- 75 m

4√mn

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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 ……………(i)

In rt. ∆ ABD

AB = BD cot 60^{°}

AB = (BC + CD) cot 60^{°}

AB = (10 + CD) * \\frac{1}{√3}\\)……………(ii)

Equating (i) and (ii)

(10 + CD) * \\frac{1}{√3}\\) = 100

(10 + CD)= 100√3

CD = 100√3 – 100

= 10(1.732 – 1) = 100 * 0.732 = 73.2 m

The question is **"To find the height of the student"**

Choice B is the correct answer.

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