The sum of the interior angles of any triangle is always 180º:
[tex]A+B+C=180[/tex]
Use the equation above and the given data to find C:
[tex]\begin{gathered} C=180º-A-B \\ C=180º-38º-72º \\ C=70º \end{gathered}[/tex]
Law of sines:
[tex]\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}[/tex]
Use the pair of ratios for a and b to solve a:
[tex]\begin{gathered} \frac{a}{sinA}=\frac{b}{sinB} \\ \\ a=sinA*\frac{b}{sinB} \\ \\ a=sin38º*\frac{12}{sin72º} \\ \\ a=7.8 \end{gathered}[/tex]
Use the pair of ratios for b and c to solve c:
[tex]\begin{gathered} \frac{c}{sinC}=\frac{b}{sinB} \\ \\ c=sinC*\frac{b}{sinB} \\ \\ c=sin70º*\frac{12}{sin72º} \\ \\ c=11.9 \end{gathered}[/tex]Thenm, the solution for the given triangle is:A=38ºB=72ºC=70ºa=7.8b=12c=11.9
