Question

if b+c=2acos(B-C)/2 then proove that <A=60°...

solve it with full steps.

Answer 1

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b+c=2acos(B-C)

ksinB+ksinC=2×ksinAcos(B-C)/2

=>k2sin(B+C)/2×cos(B-C)/2=k2SinAcos(B-C)/2

(k and 2 is cancelled here )

=>sin(B+C)/2=SinA

=>sin(B+C)/2=SinA

(sin is cancelled here )

=(B+C)=2A

=>B+C+A=A+2A {adding angle A on both side ,we get }

180°=3A { As We know that the sum of all angle of triangle is 180°}

so....3A=189°
.A=60° Prooved here .