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Question:

In the trapezium ABCD the sides AB and CD are parallel. The value of $\frac{\sin \angle BAC}{\sin \angle BAD}$ is

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Explanation:

Refer the diagram below:

In ∆ACL,
Sin∠BAC = Sin∠LAC = perpendicular/hypotenuse = CL/AC   …(1)

Similarly, in ∆ADM,
Sin∠BAD = Sin∠MAD = DM/AD   …(2)

(1) ÷ (2)

⇒ Sin∠BAC : Sin∠BAD = CL/AC : DM/AD

Now, CL = DM

Hence, Sin∠BAC : Sin∠BAD = 1/AC : 1/AD = AD : AC

Hence, option (a).