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UPSC NDA I and II 2020 Math Solved Paper

DIRECTIONS (Qs. 36-37) : Read the following information and answer the two items that follow:

A B C D

is a trapezium such that

A B

and

C D

are parallel and

B C

is perpendicular to them. Let

\angle A D B = \theta, \angle A B D = \alpha, B C = p

and

C D = q

and

C D = q

[NDA-I 2020]

Question : 37 of 120

Marks: +1, -0

What is

A B

Solution: 👈: Video Solution

(a) From

\triangle ABD

\frac{AB}{\sin\theta}=\frac{BD}{\sin(180^{\circ}-(\alpha+\theta))}

\Rightarrow AB = \frac{BD \cdot \sin\theta}{\sin(\alpha+\theta)} = \frac{BD \cdot \sin\theta}{\sin\alpha \cos\theta + \cos\alpha \sin\theta}

Again from

\triangle BCD

BD = \sqrt{(BC)^{2}+(CD)^{2}} = \sqrt{p^{2}+q^{2}}

\sin\alpha = \frac{BC}{BD} = \frac{p}{\sqrt{p^{2}+q^{2}}}

\cos\alpha = \frac{CD}{BD} = \frac{q}{\sqrt{p^{2}+q^{2}}}

\therefore AB = \frac{\sqrt{p^{2}+q^{2}} \cdot \sin\theta}{\frac{p\cos\theta}{\sqrt{p^{2}+q^{2}}} + \frac{q\sin\theta}{\sqrt{p^{2}+q^{2}}}}

= \frac{(p^{2}+q^{2}) \sin\theta}{p \cos\theta + q \sin\theta}

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