Solve right triangle ABC for all missing parts. Express angles in decimal degrees.a =100.8 km, c=385.6 km We must find A, B, b

Given:
a = 100.8
c = 385.6
For angle "C"
It's a right angle triangle so
C = 90
For angle A is:
[tex]\begin{gathered} b=\text{ base} \\ c=\text{ hypotenuse} \\ a=\text{ perpendicular} \end{gathered}[/tex][tex]\begin{gathered} \sin A=\frac{\text{ perpendicular}}{\text{ hypotenuse}} \\ \sin A=\frac{a}{c} \\ \sin A=\frac{100.8}{385.6} \\ \sin A=0.2614 \\ A=\sin ^{-1}(0.2614) \\ A=15.1537 \end{gathered}[/tex]For Angle B is:
[tex]\begin{gathered} a=\text{ base} \\ c=\text{ hypotenuse} \\ b=\text{perpendicular} \end{gathered}[/tex][tex]\begin{gathered} \cos B=\frac{\text{ base}}{\text{ hypotenuse}} \\ \cos B=\frac{a}{c} \\ \cos B=\frac{100.8}{385.6} \\ \cos B=0.2614 \\ B=\cos ^{-1}(0.2614) \\ B=74.8462 \end{gathered}[/tex]For "b" use pythagoras theorem.
[tex]\begin{gathered} (\text{hypotenuse)}^2=(\text{base)}^2+(\text{perpendicular)}^2 \\ c^2=b^2+a^2 \\ (385.6)^2=b^2+(100.8)^2 \\ 148687.36=b^2+10160.64 \\ b^2=148687.36-10160.64 \\ b^2=138526.72 \\ b=\sqrt[]{138526.72} \\ b=372.1917 \end{gathered}[/tex]