In ∆ ABC, b+c12=c+a13=a+b15 Prove that cos A2=cos B7=cos C11 - Maths -

Question

In ∆   A B C ,   b + c 12 = c + a 13 = a + b 15
Prove that cos   A 2 = cos   B 7 = cos   C 11

Global Expert · Accepted Answer

Let b+c12=c+a13=a+b15=k⇒b+c=12k, c+a=13k, a+b=15k⇒b+c+c+a+a+b=12k+13k+15k⇒2a+b+c=40k⇒a+c+b=20k⇒a+12k=20k    ∵b+c=12k⇒a=8kAlso, c+a=13k⇒c=13k-a=13k-8k=5kand a+b=15k⇒b=15k-a=15k-8k=7kNow,cosA=b2+c2-a22bc=k249+25-64k22×35=17cosB=a2+c2-b22ac=k264+25-492×40k2=12cosC=a2+b2-c22ab=k264+49-252×56k2=1114∴  cosA:cosB:cosC=17:12:1114=2:7:11⇒ cosA2=cosB7=cosC11

Hence proved

In ∆ A B C , b + c 12 = c + a 13 = a + b 15 . Prove that cos A 2 = cos B 7 = cos C 11 .

In $∆ A B C, \frac{b + c}{12} = \frac{c + a}{13} = \frac{a + b}{15}$ . Prove that $\frac{\cos A}{2} = \frac{\cos B}{7} = \frac{\cos C}{11}$ .