Could you help me approach this problem using sine law? 

Given that ABC is an isosceles triangle, $[BD]$ is angle bisector, $\angle BDA = 120^\circ$. Evaluate the degree of $\angle A $

Could you help me approach this problem using sine law? 
Here's my attempt: 
From the angle bisector theorem, we know that
$$\dfrac{|AB|}{|BC|} = \dfrac{|AD|}{|DC|} $$
In $\triangle ADB$, let's call same angles $\alpha$ and we have that
$$\dfrac{|AB|}{\sin (120)} = \dfrac{|AD|}{\sin(\alpha )} \implies \dfrac{|AB|}{|AD|} = \dfrac{\sin (120)}{\sin (\alpha )}  $$
This also equals 
$$\dfrac{|AB|}{|AD|} = \dfrac{\sin (120)}{\sin (\alpha )} = \dfrac{|BC|}{|DC|}$$
Now $\angle A 180-120-\alpha  = 60-\alpha $, then
$$\dfrac{|DB|}{\sin (x)} = \dfrac{|AB|}{\sin (120)} \implies \dfrac{|DB|}{|AB|} = \dfrac{\sin(60-\alpha )}{\sin (120)} $$
 A: I think Law of Sines is ill-suited for this problem. However, in order to understand this, you first need to see the answer. First, $AB=AC$ since triangle $ABC$ is isosceles. Therefore:
$$\angle ABC=\angle ACB$$
Also, $BC$ is an angle bisector, so:
$$\angle ABD=\frac 1 2\angle ABC$$
From triangle $ABD$, we have:
$$\angle A+\frac 1 2\angle ABC+120^\circ=180^\circ$$
From triangle $ABC$, we have:
$$\angle A+2\angle ABC=180^\circ$$
Subtract the second equation by the first:
$$\frac 3 2\angle ABC-120^\circ=0\rightarrow \angle ABC=80^\circ$$
Substitute back into the second equation:
$$\angle A+160^\circ=180^\circ\rightarrow\angle A=20^\circ$$
Thus, our final answer is $20^\circ$. This means:
$$\sin A=\frac i 2\left(\sqrt[3]{\frac{-1-\sqrt{-3}}{2}}-\sqrt[3]{\frac{-1+\sqrt{-3}}{2}}\right)$$
I think it would be very hard to derive this complex expression from using Law of Sines in order to solve for A, which is why Law of Sines is not a good way to solve this rather simple angle problem.
A: Let $x$ be the measure of $\angle ABD$
and $y$ be the measure of $\angle BAC$
$x+y+ 120 = 180\\
4x + y = 180$
And solve the system of equations.
Regrading law of sines.. that might be useful you knew more side lengths.  Right now all you know is that the triangle is isosceles.
A: We have that since $\angle BDC=60° \implies \angle ACB=\angle ABC=80*$ and $\angle BAC=20°$.
To find the result by law of sine let


*

*$BC=a $

*$AB=AC=b$

*$BD=x$

*$DC=y$

*$\angle BAC=\alpha$
then we have to solve the following systems of $5$ equations in $5$ unknowns


*

*$\frac{a}{\sin \alpha}=\frac{b}{\sin 80}$

*$\frac{a}{\sin 60}=\frac{x}{\sin 80}=\frac{y}{\sin 40}$

*$\frac{x}{\sin \alpha}=\frac{b}{\sin 120}=\frac{b-y}{\sin 40}$
