The sum of all angles in any triangles is \(\displaystyle{180}^{\circ}\), so

\(\displaystyle\angle{a}+\angle{b}+\angle{c}={180}\)

Substitute \(\displaystyle\angle{c}=\angle{b}{\quad\text{and}\quad}\angle{a}={4}\angle{b}\) to solve for \(\displaystyle\angle{b}\):

\(\displaystyle{4}\angle{b}+\angle{b}+\angle{b}={180}^{\circ}\)

\(\displaystyle{6}\angle{b}={180}^{\circ}\)

\(\displaystyle\angle{b}=\frac{{{180}^{\circ}}}{{6}}\)

\(\displaystyle\angle{b}={30}^{\circ}\)

Substitute \(\displaystyle\angle{c}=\angle{b}\) to solve for \(\displaystyle\angle{c}\):

\(\displaystyle\angle{c}={30}^{\circ}\)

Substitute \(\displaystyle\angle{a}={4}\angle{b}\to\) solve for \(\displaystyle\angle{a}\):

\(\displaystyle\angle{a}={4}\times{30}^{\circ}\)

\(\displaystyle\angle{a}={120}^{\circ}\)

The answer:

\(\displaystyle\angle{b}={30}^{\circ},\angle{c}={30}^{\circ},\angle{a}={120}^{\circ},\)

\(\displaystyle\angle{a}+\angle{b}+\angle{c}={180}\)

Substitute \(\displaystyle\angle{c}=\angle{b}{\quad\text{and}\quad}\angle{a}={4}\angle{b}\) to solve for \(\displaystyle\angle{b}\):

\(\displaystyle{4}\angle{b}+\angle{b}+\angle{b}={180}^{\circ}\)

\(\displaystyle{6}\angle{b}={180}^{\circ}\)

\(\displaystyle\angle{b}=\frac{{{180}^{\circ}}}{{6}}\)

\(\displaystyle\angle{b}={30}^{\circ}\)

Substitute \(\displaystyle\angle{c}=\angle{b}\) to solve for \(\displaystyle\angle{c}\):

\(\displaystyle\angle{c}={30}^{\circ}\)

Substitute \(\displaystyle\angle{a}={4}\angle{b}\to\) solve for \(\displaystyle\angle{a}\):

\(\displaystyle\angle{a}={4}\times{30}^{\circ}\)

\(\displaystyle\angle{a}={120}^{\circ}\)

The answer:

\(\displaystyle\angle{b}={30}^{\circ},\angle{c}={30}^{\circ},\angle{a}={120}^{\circ},\)