Sine rule exam question (with $\sin 2 \theta$ term)

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Triangle diagram In the triangle ABC, AC = 3cm, BC = 2cm, $\angle$ BAC = $\theta$ and $\angle$ ABC = 2$\theta$. Calculate the value of $\theta$ correct to the nearest tenth of a degree.

The above is part of an A-Level exam question. I am assuming you use the sine rule to get $\theta$, but I am not sure how to extract this value. The answer is given as 41.4$^o$.

I have looked at the current proofs and rules given in textbook, but get stuck at:

$$\frac{\sin \theta}{2}=\frac{\sin 2\theta}{3}$$

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Your attempt is fine. But then you need to know the formula $$\sin2\theta=2\sin\theta\cos\theta$$ After that it is easy. You get $\cos\theta=3/4$. Since $\theta,2\theta$ are angles in a triangle, there sum is less than $180^o$, so $\theta<60^o$ and you get $\theta=41.4^o$.