help with trigonometry problem

**yiapanisa** · May 21st 2011, 01:22 PM

Triangle ABΓ if α+γ=2β and Γ=120ο show that α/3=β/5=γ/7
please help
thanks
alex

**Opalg** · May 22nd 2011, 06:56 AM

Originally Posted by yiapanisa

Triangle ABΓ: if α+γ=2β and Γ=120º, show that α/3=β/5=γ/7

By the cosine rule, $\gamma^2 = \alpha^2+\beta^2- 2\alpha\beta\cos\Gamma = \alpha^2+\beta^2 + \alpha\beta$ , since $\cos(120^\circ) = -\tfrac12$ . Thus $\alpha^2-\gamma^2 + \alpha\beta + \beta^2 = 0$ . But $\alpha^2-\gamma^2 = (\alpha+\gamma)(\alpha-\gamma) = 2\beta(\alpha-\gamma)$ . Deduce that $3\alpha+\beta -2\gamma = 0$ . Combine that with the given equation $\alpha-2\beta +\gamma = 0$ to conclude that $\alpha:\beta:\gamma$ are in the ratio $3:5:7$ .

**yiapanisa** · May 22nd 2011, 10:07 AM

thx a lot

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