The sum of the angle measures of a triangle is 180° so we can write:
\(\displaystyle{m}∠{A}+{m}∠{B}+{m}∠{C}={180}°\)

Substitute \(\displaystyle{m}∠={30}°{\quad\text{and}\quad}{P}{S}{K}{m}∠{A}={12}{m}∠{C}\): \(\displaystyle\frac{{1}}{{2}}{m}{<}{C}+{30}°+{m}{<}{C}={180}°\)</span>

Solve for \(\displaystyle{m}∠{C}\)

\(\displaystyle\frac{{3}}{{2}}{m}{<}{C}+{30}°={180}°\)</span>

\(\displaystyle\frac{{3}}{{2}}{m}{<}{C}={150}°\)</span>

\(\displaystyle{m}{<}{C}=\frac{{2}}{{3}}\cdot{150}°\)</span>

\(\displaystyle{m}{<}{C}={100}°\)</span>

Hence, \(\displaystyle{m}{<}{A}=\frac{{1}}{{2}}{\left({100}°\right)}\)</span>

PSKm<A=50°

Substitute \(\displaystyle{m}∠={30}°{\quad\text{and}\quad}{P}{S}{K}{m}∠{A}={12}{m}∠{C}\): \(\displaystyle\frac{{1}}{{2}}{m}{<}{C}+{30}°+{m}{<}{C}={180}°\)</span>

Solve for \(\displaystyle{m}∠{C}\)

\(\displaystyle\frac{{3}}{{2}}{m}{<}{C}+{30}°={180}°\)</span>

\(\displaystyle\frac{{3}}{{2}}{m}{<}{C}={150}°\)</span>

\(\displaystyle{m}{<}{C}=\frac{{2}}{{3}}\cdot{150}°\)</span>

\(\displaystyle{m}{<}{C}={100}°\)</span>

Hence, \(\displaystyle{m}{<}{A}=\frac{{1}}{{2}}{\left({100}°\right)}\)</span>

PSKm<A=50°