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B = pi/3, C = pi/4, D divides BC in 1:3: find sin(BAD)/sin(CAD)

The same question, asked 2 times: 12 May 2024 S2, 15 Apr 2026 S2. Every wording below, oldest first.

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  1. Question 1

    12 May 2024 S2

    129
    In △ABC\triangle ABC, m∠B=π3m\angle B = \frac{\pi}{3} and m∠C=π4m\angle C = \frac{\pi}{4}. Let point D divide BC internally in the ratio 1:31:3, then sin⁡(∠BAD)sin⁡(∠CAD)\frac{\sin(\angle BAD)}{\sin(\angle CAD)} has the value

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  2. Question 2

    15 Apr 2026 S2

    65
    In a triangle ABC, with the usual notations, ∠B=π3,∠C=π4\angle B = \dfrac{\pi}{3}, \angle C = \dfrac{\pi}{4}. If D divides BC internally in the ratio 1:3, then sin⁡∠BADsin⁡∠CAD=\dfrac{\sin\angle BAD}{\sin\angle CAD} =

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