a=8. c=5, B=40 How do I get A with sin B/b=sinA/a

You're correct Guest.....I did everything wrong....LOL!!!....let me try again!!!

Using  [the  correct !!!] Law of Cosines to find b, we have

b^2 = 8^2 + 5^2 - 2(8)(5)cos(40)

b =  sqrt [8^2 + 5^2 - 40cos(40)]  = about 5.26

So we have

SinA / a = SinB/ b 

SinA / 8  = sin(40) / 5.26

SinA  = 8*sin(40)/5.26

And using the Sine inverse, we have

Sin-1 [8* sin (40) / 5.26 ]  = A = about 78.37°

[Thanks, Guest for spotting my stupid mistakes  !!!  ]

^{..............}

Sorry Chris.I think you are doing soemthing wrong

first of all,one of the law of cosine is 

\(b^2=a^2+c^2-2ac*cosB\)

not

\(b^2=a^2+c^2-2ac*sinB\)

and this not a SSA situtiation.note that the given is side a,side c and angle B

So this a SAS situation.