Nú er $\alpha+\beta+\gamma=180^\circ$ og þar sem gefið er að $2\beta=\alpha+\gamma$ þá fæst að $2\beta=180^\circ-\beta$ og því $\beta=60^\circ$. Þá er $\alpha+\gamma=120^\circ$ og því $$ \begin{aligned} \cos(\gamma)&=\cos(120^\circ-\alpha)=\cos(120^\circ) \cos(\alpha)+\sin(120^\circ) \sin(\alpha)\\&=-\frac{1}{2}\cos(\alpha)+\frac{\sqrt3}{2}\sin(\alpha),\\ \sin(\gamma)&=\sin(120^\circ-\alpha)=\sin(120^\circ) \cos(\alpha)-\cos(120^\circ) \sin(\alpha)\\ &=\frac{\sqrt3}{2}\cos(\alpha)+\frac{1}{2}\sin(\alpha). \end{aligned} $$ Þá er $$ \begin{aligned} \frac{\sin(\alpha)+\sin(\beta)+\sin(\gamma)}{\cos(\alpha)+\cos(\beta)+\cos(\gamma)} &=\frac{\sin(\alpha)+\frac{\sqrt3}{2}+\frac{\sqrt{3}}{2}\cos(\alpha) +\frac{1}{2}\sin(\alpha)}{\cos(\alpha)+\frac{1}{2}-\frac{1}{2}\cos(\alpha)+\frac{\sqrt3}{2}\sin(\alpha)}\\ &=\frac{3\sin(\alpha)+\sqrt{3}+\sqrt{3}\cos(\alpha)}{\cos(\alpha)+1+\sqrt{3}\sin(\alpha)}\\ &=\frac{\sqrt{3}(\sqrt{3}\sin(\alpha)+1+\cos(\alpha))}{\cos(\alpha)+1+\sqrt{3}\sin(\alpha)}\\ &=\sqrt{3} \end{aligned} $$