# Find base in isosceles triangle

If angle alpha and side $b$ is known in this isosceles triangle, how long is the base $a$? I know this is very basic but I don't know any trigonometry so I don't really know what to do here.

$$\frac{a}{\sin\alpha} = \frac{b}{\sin(\frac{180 - \alpha}{2})}$$
$$a = \frac{b\cdot \sin\alpha}{\sin(\frac{180 - \alpha}{2})}$$
hint: $a^2 = b^2+b^2-2b^2\cos(\alpha)$ is known as the law of cosine in a triangle. Can you take it from here?