#### Answer

We can draw one triangle with the given parts.
$A = 40^{\circ}, B = 33^{\circ},$ and $C = 107^{\circ}$

#### Work Step by Step

We can use the law of sines to find the angle $B$:
$\frac{a}{sin~A} = \frac{b}{sin~B}$
$sin~B = \frac{b~sin~A}{a}$
$sin~B = \frac{(30)~sin~(40^{\circ})}{35}$
$B = arcsin(0.551)$
$B = 33^{\circ}$
We can find angle $C$:
$A+B+C = 180^{\circ}$
$C = 180^{\circ}-A-B$
$C = 180^{\circ}-40^{\circ}-33^{\circ}$
$C = 107^{\circ}$
Note that we can also find another angle for B.
$B = 180-33^{\circ} = 147^{\circ}$
However, we can not form a triangle with this angle B and angle A since these two angles sum to more than $180^{\circ}$