<s>[INST] Find an angle of a triangle on a larger triangle which cut through its midpoint 
In triangle $\triangle BAC$ with $\angle ABC = 30\deg$. $D$ is the midpoint of $BC$. We join $A$ and $D$ and $\angle CDA = 45 \deg$. Find $\angle BAC$.


On applying Sine rule,
$$\frac{2x}{\sin {(15+\theta)}}=\frac{AC}{\sin 30}$$
and also
$$\frac{x}{\sin \theta}=\frac{AC}{\sin 45}$$
Where $x$ is $CD$ or $DB$ and $\theta$ is $\angle CAD$.
But solving this gives $$\frac{\sin {(15+\theta)}}{\sin \theta}=\sqrt 2$$
Is this correct?
 [/INST] Your reasoning looks good to me. Using your second equation, $$AC=\frac{x}{\sqrt{2}\sin{\theta}}$$ Now substituting $AC$ in the first equation, $$\frac{2x}{\sin{(15+\theta)}}=\frac{2x}{\sqrt{2}\sin{\theta}}$$ or $$\sin{(15+\theta)}=\sqrt{2}\sin{\theta}$$
Using trig identity, $$\cos15\sin\theta+\sin15\cos\theta=\sqrt{2}\sin{\theta}$$ Dividing by $\sin\theta$ we get $$\cot\theta=\frac{\sqrt{2}-\cos15}{\sin15}$$ Knowing that $\sin15=\frac{\sqrt{3}-1}{2\sqrt{2}}$, $\cos15=\frac{\sqrt{3}+1}{2\sqrt{2}}$ we get $\cot\theta=\sqrt{3}$,  $\theta=30°$
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