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?
 A: you will Need three equations
$$a^2=b^2+c^2-2bc\cos(\alpha)$$
$$a=\frac{b\sin(\alpha)}{\sin(30^{\circ})}$$
$$c=\frac{b\sin(150^{\circ})}{\sin(30^{\circ})}$$
then you Can divide by $b^2$ and you will get only an equation for $$\alpha$$
A: 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 $\cot\theta=\sqrt{3}$,  $\theta=30°$
A: Never Underestimate the Power of Euclidean Geometry



*

*By Exterior Angle Theorem $\angle DAB = 15°$.

*Draw from $C$ the line perpendicular to $AB$, intersecting $AB$ in $E$. We then have $\angle ECB = 60°$ and thus $\triangle EBC$ is half of an equilater triangle and $CE \cong \frac{BC}{2} \cong CD \cong BD$.

*Now connect $E$ with $D$. $\triangle CED$ is isosceles, but having $\angle ECB = 60°$, it is also equilateral and thus $ED \cong CE$, and $\angle EDA = \angle DAB = 15°$.

*So $\triangle AED$ is isosceles and we get also $AE \cong CE$.

*We conclude that $\triangle ACE$ is isosceles and right-angled. Therefore $\angle BAC = 45°$.


$\blacksquare$
