In Fig : what is $\angle DAC= ?$

Question

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?

Answer 1

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 $y$.

Answer 2

$E$ is the intersection between the line through $CD$ and $AB$.

It means that $∠BDE=60°→∠BED=90°$. Let's call $AB=AC=l$ and $BD=2p$.

Considering the right triangle $△AEC$ we have $AE=l\sin10°$.

Considering the right triangle $△BED$ we have that $BE=p\sqrt{3}$ and $ED=p$. Then

$$BE=p\sqrt{3}=AB−AE=l−l\sin10°→p=\frac{l(1−\sin10°)}{\sqrt{3}}$$

$$\tan(∠EAD)=\tan(80°−x)=\frac{ED}{AE}=\frac{1−\sin10°}{\sqrt{3}\sin10°}≈2.7475=\tan(70°)$$

I used Wolfram for the last equality, so $80°−x=70°→x=10°80°−x=70°→x=10°$.