two sides and angle between them triangle question.

How about the law of cosines?

Consider the following triangle $\triangle ABC$, the $ \color{maroon} {\text{poly 1}}$ below, with sides $\color{maroon}{\overline{AB}=c}$ and $\color{maroon}{\overline{AC}=b}$ known. Further the angle between them, $\color{green}\alpha$ is known.

$\hskip{2 in}$

Then, the law of cosines tell you that $$\color{maroon}{a^2=b^2+c^2-2bc\;\cos }\color{blue}{\alpha}$$

This is precisely what the cosine theorem allows you to compute: $$c^2 = a^2 + b^2 - 2ab\cos(\gamma)$$ where $\gamma$ is the angle opposite to $c$

(uhm, yes, is probably be called 'law of cosine' rather than 'cosine theorem').

Use the law of cosines...

$c^2 = a^2 + b^2 - 2ab \cdot \cos{\theta}$

... where $a$ and $b$ are the sides you know, $\theta$ the angle between them, and $c$ the side you seek, opposite $\theta$.