Friday, January 15, 2016

geometry - 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?



Answer



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.


No comments:

Post a Comment

analysis - Injection, making bijection

I have injection $f \colon A \rightarrow B$ and I want to get bijection. Can I just resting codomain to $f(A)$? I know that every function i...