How would I find sin B, Cos B, sin C, and Cos C

The given triangle is
We get hypotenuse =BC=40 by observing the given triangle.
Consider the angle B.
For angle B, the Opposite side is AC=32 adjacent side AB =24.
We know that
[tex]\sin \theta=\frac{Opposite\text{ side}}{\text{Hyponetuse}}[/tex][tex]\sin B=\frac{AC}{BC}[/tex]Substitute AC=32 and BC=40, we get
[tex]\sin B=\frac{32}{40}=0.8[/tex]We know that
[tex]\cos \theta=\frac{adjacent\text{ side}}{\text{Hypotenuse}}[/tex][tex]\cos B=\frac{AB}{BC}[/tex]Substitute AB=24 and BC=40, we get
[tex]\cos B=\frac{24}{40}=0.6[/tex]Consider the angle C.
For angle C, the Opposite side is AB=24 and the adjacent side AC=32.
[tex]\sin C=\frac{AB}{BC}[/tex]Substitute AB=24 and BC=40, we get
[tex]\sin C=\frac{24}{40}=0.6[/tex][tex]\cos C=\frac{AC}{BC}[/tex]Substitute AC=32 and BC=40, we get
[tex]\cos C=\frac{32}{40}=0.8[/tex]Hence the answers are
[tex]\sin B=0.8[/tex][tex]\cos B=0.6[/tex][tex]\sin C=0.6[/tex][tex]\cos C=0.8[/tex]