Triangle ABΓ if α+γ=2β and Γ=120ο show that α/3=β/5=γ/7
please help
thanks
alex
By the cosine rule, $\displaystyle \gamma^2 = \alpha^2+\beta^2- 2\alpha\beta\cos\Gamma = \alpha^2+\beta^2 + \alpha\beta$, since $\displaystyle \cos(120^\circ) = -\tfrac12$. Thus $\displaystyle \alpha^2-\gamma^2 + \alpha\beta + \beta^2 = 0$. But $\displaystyle \alpha^2-\gamma^2 = (\alpha+\gamma)(\alpha-\gamma) = 2\beta(\alpha-\gamma)$. Deduce that $\displaystyle 3\alpha+\beta -2\gamma = 0$. Combine that with the given equation $\displaystyle \alpha-2\beta +\gamma = 0$ to conclude that $\displaystyle \alpha:\beta:\gamma$ are in the ratio $\displaystyle 3:5:7$.