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:

https://en.wikibooks.org/wiki/Trigonometry/Solving_triangles_by_half-angle_formulae ]

the derivation is pretty easy. Please let me know if you understood.

  • $\begingroup$ 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! $\endgroup$ – 674123173797 - 4 May 28 '18 at 18:48

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