Answer
See below
Work Step by Step
We are given $a, c, b$. Use law of cosines to find $c$:
$$a^2=a^2+a^2-2aa\cos A\\a^2=2a^2-2a^2\cos A\\\cos A=\frac{1}{2}$$
Doing the same, we get: $\cos B=\frac{1}{2}\\\cos C=\frac{1}{2}$
Find: $\arccos(\cos A)=\arccos (\frac{1}{2})\\A=\arccos (\frac{1}{2})\\ \rightarrow A_1=60^\circ \vee A_2=300^\circ\\\rightarrow A=60^\circ$
Since the sum of the triangle is $180^\circ$, we obtain:
$$A+B+C=180^\circ\\60^\circ+60^\circ+60^\circ=180^\circ$$
The following equality is true. All the angles in an equilateral triangle have the measure of $60^\circ$.
