If △ABC,8R2=a2+b2+c2, then the triangle ABC, is
right angled
isosceles
equilateral
none of these
We have,
8R2=a2+b2+c2⇒8R2=4R2sin2A+sin2B+sin2C⇒sin2A+sin2B+sin2C=2⇒1−cos2A+1−cos2B+sin2C=2⇒cos2A−sin2C+cos2B=0⇒cos(A+C)cos(A−C)+cos2B=0⇒−cosB{cos(A−C)−cosB}=0⇒−cosB{cos(A−C)+cos(A+C)}=0⇒−2cosAcosBcosC=0⇒cosA=0 or, cosB=0 or, cosC=0⇒A=π2 or, B=π2 or, C=π2
⇒ △ABC is a right angled triangle