Question

In any triangle âabc, show that â¬(b+c)cos a = 2s where s = (a+b+c)/2

Answer 1

a=bcosC+ccosB

b=acosC+ccosA

c=bcosA+acosB

adding together we get

(a+b)cosC+(b+c)cosA+(c+a)cosB=a+b+c