(CosA+2CosC)/(CosA+2CosB)= SinB/SinC, Prove that the triangle is either isosceles or right angled?

1 Answer
Mar 4, 2018

Given rarr(cosA+2cosC)/(cosA+2cosB)=sinB/sinC

rarrcosAsinB+2sinB*cosB=cosAsinC+2sinCcosC

rarrcosAsinB+sin2B=cosAsinC+sin2C

rarrcosA(sinB-sinC)+sin2B-sin2C=0

rarrcosA[2sin((B-C)/2)*cos((B+C)/2)]+2*sin((2B-2C)/2)*cos((2B+2C)/2)]=0

rarrcosA[2sin((B-C)/2)*cos((B+C)/2)]+2*sin(B-C)*cos(B+C)]=0

rarrcosA[2sin((B-C)/2)*cos((B+C)/2)]+cosA*2*2*sin((B-C)/2)*cos((B-C)/2)]=0

rarr2cosA*sin((B-C)/2)[cos((B+C)/2)+2cos((B-C)/2)]=0

Either, cosA=0 rarrA=90^@

or, sin((B-C)/2)=0 rarrB=C

Hence, the triangle is either isosceles or right angled. Credit goes to dk_ch sir.