[tex]sinP=\frac{Perpendicular}{hypotenuse} OR \frac{Opposite}{hypotenuse}[/tex]
Use the Pythagoras' (Pythagorean) Theorem:-
[tex]hyp=\sqrt{5^{2} +12^{2} }[/tex]
[tex]hyp=13[/tex]
Substitute 5 and 13 in sinP:-
[tex]sinP=\frac{5}{13}[/tex]
[tex]sinP=0.38[/tex]
