The values you've given for  and B are unrealistic.

There is no triangle with such values for angles.

In any case, use the formular \frac{a}{SinA} = \frac{b}{SinB} = \frac{c}{SinC}

You're asked to look for c, so let's get the angle C.

C = 180 - (A + B)

Then, use \frac{a}{SinA} = \frac{c}{SinC}

Making c the subject of the formular;
c = \frac{aSinC}{SinA}

Hope that was helpful?
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