Have you covered the cosine rule?
$\displaystyle c^2=a^2+b^2-2ab\cos C$ which may also be written as $\displaystyle \cos C = \dfrac{a^2+b^2-c^2}{2ab}$
Edit: Misread angle as a side - use the law of sines instead (post left here for continuity)
My apologies, I misread that angle as a side. Instead you can use the sine rule
$\displaystyle \dfrac{\sin A}{a} = \dfrac{\sin B}{b}$
In this case you have
- $\displaystyle a = 6$
- $\displaystyle \sin B = 96$
- $\displaystyle b = 10$
$\displaystyle \sin A = \dfrac{a \sin B}{b} = \dfrac{6 \sin(96)}{10}$