Solution:
Given,
We know, the sum of all the interior angles of a triangle is .
Thus,
\begin{array}{l}
9 \angle A=180^{\circ} \\
\angle A=20^{\circ}
\end{array}
\begin{array}{l}
20^{\circ}+4 \angle B=180^{\circ} \\
\angle B=40^{\circ} \\
3 \angle B=\angle C \\
\angle C=3 \times 40=120^{\circ}
\end{array}
\begin{array}{l}
\angle B=40^{\circ} \\
\angle C=120^{\circ}
\end{array}