Let the measure of angle A is x
then the measure of B
[tex]B=\frac{3}{5}(180-x)[/tex]The measure of C is
[tex]\begin{gathered} C=\frac{4}{3}(90-\frac{3}{5}(180-x)) \\ =120-\frac{4}{5}(180-x) \\ =120-144+\frac{4}{5}x \end{gathered}[/tex]Now
[tex]\begin{gathered} x+(108-\frac{3}{5}x)+(\frac{4}{5}x-24)=180 \\ x-\frac{3}{5}x+\frac{4}{5}x+84=180 \\ \frac{6}{5}x=96 \\ x=80 \end{gathered}[/tex]So
[tex]\begin{gathered} \angle A=80 \\ \angle B=\frac{3}{5}\times100=60 \\ \angle C=\frac{4}{3}\times30=40 \end{gathered}[/tex]