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In a triangle ABC,∠B = π/3 and ∠ π/3. Let D divide BC internally in the ratio 1:3. Then sin∠BAD/sin∠CAD equals

(A)  1/√6

(B)   1/3

(C)  1/√3

(D)   √(2/3)

1 Answer

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by (53.8k points)
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Best answer

Correct option  (A)  1/√6

See Fig.

BD : DC = 1 : 3,

To calculate : sin∠BAD/sin∠CAD

Apply sine rule

In ∆ABD,

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