Got it...
so B and b are given
you have this:
sin(17)/10 = Sin(A)/a = Sin(B)/19
you can bring 19 over to find to the left
so u got the left hand side = sin(B)
now just evaluate that, and thats = to Sin(B) so take the arcsin
u got B
A will be whatever degrees you have left. You can play with those values
and get similar triangles as long as they agree with equality above. Sorry i made this overly complicated at first : /
so B and b are given
you have this:
sin(17)/10 = Sin(A)/a = Sin(B)/19
you can bring 19 over to find to the left
so u got the left hand side = sin(B)
now just evaluate that, and thats = to Sin(B) so take the arcsin
u got B
A will be whatever degrees you have left. You can play with those values
and get similar triangles as long as they agree with equality above. Sorry i made this overly complicated at first : /