Given the shown triangle
Using the pithagorean theorem
[tex]AB^2=9^2+12^2[/tex][tex]AB=\sqrt{81+144}[/tex][tex]AB=15[/tex]
AB=15
Using law of sines
[tex]\frac{a}{sin(A)}=\frac{b}{Sin(B)}=\frac{c}{sin(C)}[/tex][tex]\frac{12}{sin(A)}=\frac{9}{Sin(B)}=\frac{15}{sin(90)}[/tex]
Solving for A
[tex]\frac{12}{Sin(A)}=\frac{15}{sin(90)}[/tex][tex]\frac{12}{15}=Sin(A)[/tex][tex]A=ArcSin(\frac{4}{5})[/tex][tex]A=53.13[/tex]
A=53.13°
Solving for B
[tex]\frac{9}{S\imaginaryI n(B)}=\frac{15}{s\imaginaryI n(90)}[/tex][tex]\frac{9}{15}=Sin(B)[/tex][tex]B=ArcSin(\frac{3}{5})[/tex][tex]B=36.87[/tex]
B=36.87°
