Consider the following: 1. AD θ = AB α 2. BD θ = AB (θ + α) Which of the above is / are correct?

Correct Answer is : Both 1 and 2

Correct Answer is : Both 1 and 2

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UPSC NDA I and II 2020 Math Solved Paper

Question Numbers: 6-7

Direction: Read the following information and answer the two items that follow:

ABCD is a trapezium such that AB and CD are parallel and BC is perpendicular to them. Let ∠ ADB = θ, ∠ ABD = α, BC = p and CD = q.

Question : 6 of 120

Marks: +1, -0

Consider the following:

1. AD \sin \theta = AB \sin \alpha

2. BD \sin \theta = AB \sin (\theta + \alpha)

Which of the above is / are correct?

Solution:

Concept:

Apply the sine rule in triangle

ABD

Explanation:

Statement 1:

In triangle

ABD

, the sine rule gives:

\frac{\sin \theta}{AB} = \frac{\sin \alpha}{AD}

Cross-multiplying yields

AD \sin \theta = AB \sin \alpha

Thus, statement 1 is correct.

Statement 2:

In the same triangle, the sine rule also gives:

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

Since

\sin(180^\circ - x) = \sin x

, we have:

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

Cross-multiplying gives

BD \sin \theta = AB \sin(\theta + \alpha)

Thus, statement 2 is also correct.

Answer:

Both statements 1 and 2 are correct. Option C.

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