Find all solutions for each triangle.
A = 30
a = 4
b = 8
4 / 30 = 8 / SIN B
4 SIN B = 8 ( SIN30 )
SIN B = 8 (SIN 30) / 4 = 1
Did I do that right?
I know there is only one solution.
Yes, you are right.
$\displaystyle
=\frac{4}{sin30}= \frac{8}{sinB}
$
$\displaystyle
sinB=1
$
$\displaystyle
B=90
$
Now, in right triangle, right angled at B,
$\displaystyle
c=\sqrt{8^2-4^2}
$
$\displaystyle
c=\sqrt{48}=\;4\sqrt{3}
$
$\displaystyle
C=180-(90+30)=60
$