Given $\angle A = 60°, \angle B = 80°, \angle C = 40°$, find $\alpha$ 
I tried making an equilateral triangle with side $b$ but it didn't seem to be of any use. Please help.
 A: 

*

*Construct $BC=AC=a+b+c$

*Mark $EF=ED=c$. So $FB=b$

*$\triangle DEF$ is $100^{\circ}-40^{\circ}-40^{\circ}$ triangle, so is $\triangle CDF$

*$CD=DF=b=FB$.

*$\angle BFD=140^{\circ} \Rightarrow \angle FBD = \angle FDB = 20^{\circ}$

*In $\triangle ACD$, $$ \dfrac{b}{\sin \alpha} = \dfrac{a+b+c}{\sin 60^{\circ}}$$
In $\triangle BCD$, $$ \dfrac{b}{\sin 20^{\circ}} = \dfrac{a+b+c}{\sin 120^{\circ}}$$
We conclude $$ \boxed{\alpha = 20^{\circ}}$$

Thanks to @Nanayajitzuki, we can also finish without resorting to sine law.


*

*Construct equilateral $\triangle GCD$ with $G$ on $AD$.

*In $\triangle CGA$ and $\triangle CDB$, $\angle CGA = 120^{\circ} = \angle CDB$ . $CD=b=CG$. $CA=CB$.

*So $\triangle CGA \cong \triangle CDB$ by SSA congruence condition** $$\boxed{\therefore \alpha = 20^{\circ}}$$
Note**
The ambiguity in SSA congruence condition arises only when involved angle is acute (it leads to precisely two possible triangles as shown below $\triangle ABD$ and  $\triangle ABE$). If obtuse angle is involved, there is no ambiguity. So its use here is justified.

A: Hint : you need a calculator and the law of sines.
You can express both b and c in terms of a, via the law of sines.
then, you can use the law of sines again on the left hand triangle re 100 degrees -> (a + b + c) and $\alpha$ -> a.
