<s>[INST] In Fig : what is $\angle DAC= ?$ In Fig : what is $$\angle DAC= ? $$ 

in triangle $\triangle ABC $ And $|AB|=|AC|$And Point $D$ in the triangle $\triangle ABC $ 
such that : $\angle BCD =40,\angle CBD =20$
My Try:
$$\frac{\sin 30}{AD}=\frac{\sin \angle BAD}{BD}$$
and :
$$\frac{\sin 10}{AD}=\frac{\sin \angle DAC}{DC}$$
So :
$$\sin \angle DAC =\frac{\sin 10}{\frac{\sin 30 \times BD}{\sin \angle BAD}}\times DC$$
$$\sin \angle DAC =\frac{\sin 10 \times \sin \angle BAD}{\sin 30 \times BD}\times DC$$
Now what?
 [/INST] Let $x = \angle DAB, y = \angle DAC$. Since sum of all angles of the outer triangle is $360^\circ$, so $x + y = 80$. You can get another equation using the sine rule for the sides $BD, AD, CD$;
$$ 1 = \frac{BD}{AD} . \frac{AD}{CD} . \frac{CD}{BD} = \frac{\sin x}{\sin 30} . \frac{\sin 10}{\sin y} . \frac{\sin 20}{\sin 40}; $$
so solving the two equations $$\frac{\sin x}{\sin y} = \frac{\sin 30 \sin 40}{\sin 10 \sin 20}$$ and $$x+y = 80$$ will give you the angle $y$. 
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