# Sides of a triangle given perimeter and two angles

Let be a triangle with angles $\alpha$, $\beta$ and $\gamma.$ Let $p$ the semiperimeter of this triangle. How can I prove that the length of the opposite side to angle $\alpha$ is

$$\frac{ p\sin(\frac{\alpha}{2})}{ \cos(\frac{\beta}{2})\cos(\frac{\gamma}{2}) }$$

Using properties of area and the inradius, ($A = pr$ where $r$ is the radius of the inscribed circle and Heron's Formula $A = \sqrt{p(p-a)(p-b)(p-c)}$) I can't solve the question. How can I proceed?

If you use the formulae for $\sin\frac{\alpha}{2}$, $\cos\frac{\beta}{2}$ and $\cos\frac{\gamma}{2}$ [for example, from here:
• Yes, I understood. You just need realize that $a^2bc = \frac{s^2 (s-b)(s - c)}{\cos(\frac{B}{2}) \cdot \cos(\frac{C}{2}) }$ and use $bc = \frac{s(s-b)(s-c)}{\sin(\frac{A}{2})}$. Thanks for your help! – 674123173797 - 4 May 28 '18 at 18:48