Question

$\text{In in any}\Delta ABC,\frac{b+c}{12}=\frac{c+a}{13}=\frac{a+b}{15},\text{then prove that}\frac{cosA}{2}=\frac{cosB}{7}=\frac{cosC}{11}.$

Answer 1

$Let\frac{b+c}{12}=\frac{c+a}{13}=\frac{a+b}{15}\lambda \left(say\right)b+c=12\lambda ,c+a=13\lambda ,a+b=15\lambda (b+c+c+a+a+b)12\lambda +13\lambda +15\lambda 2(a+b+c)=40\lambda a+b+c=20\lambda b+c=12\lambda anda+b+c=20\lambda \Rightarrow a=8\lambda c+a=13\lambda anda+b+c=20\lambda \Rightarrow b=7\lambda a+b=15\lambda anda+b+c=20\lambda \Rightarrow c=5\lambda cosA=\frac{b^2+c^2-a^2}{2bc}\frac{49{\lambda }^2+25{\lambda }^2-64{\lambda }^2}{70{\lambda }^2}=\frac{1}{7}cosB=\frac{a^2+c^2-b^2}{2ac}=\frac{64{\lambda }^2+25{\lambda }^2-49{\lambda }^2}{80{\lambda }^2}=\frac{1}{2}cosC=\frac{a^2+b^2-c^2}{2ab}=\frac{64{\lambda }^2+49{\lambda }^2-25{\lambda }^2}{112{\lambda }^2}=\frac{11}{14}cosA:cosB:cosC=\frac{1}{7}:\frac{1}{2}:\frac{11}{14}=2:7:11hence\frac{cosA}{2}=\frac{cosB}{7}=\frac{cosC}{11}$