Question

In a triangle abc then the value of acosa +bcosb +ccosc/ a+b+c

Answer 1

Applying Cosine rule:

cosA=b2+c2−a22bccos⁡A=b2+c2−a22bc

⟹a.cosA=a.b2+a.c2−a32bc=a2.b2+a2.c2−a42abc⟹a.cos⁡A=a.b2+a.c2−a32bc=a2.b2+a2.c2−a42abc

Similarly, b.cosB=b2.c2+a2.b2−b42abcb.cos⁡B=b2.c2+a2.b2−b42abc

c.cosC=a2.c2+b2.c2−c42abcc.cos⁡C=a2.c2+b2.c2−c42abc

Let S=a.cosA+b.cosB+c.cosCa+b+c=S=a.cos⁡A+b.cos⁡B+c.cos⁡Ca+b+c=

a2.b2+a2.c2−a4+b2.c2+a2.b2−b4+a2.c2+b2.c2−c42abc(a+b+c)a2.b2+a2.c2−a4+b2.c2+a2.b2−b4+a2.c2+b2.c2−c42abc(a+b+c)

S=2.a2.b2+2.b2.c2+2.c2.a2−a4−b4−c42abc(a+b+c)S=2.a2.b2+2.b2.c2+2.c2.a2−a4−b4−c42abc(a+b+c)

S=−(a2+b2+c2)22abc(a+b+c)