IIT JEE , AIEEE Forum - Mathematics , Trignometry

Radon222

We have R=abc/4$ (where $ is the area of the triangle )

$=abc/4R
$=K(sinAsinBsinC).....(i) [where K=k³/4R]
$=K(sinAsinBsin(A+B))

keeping B constant and diffrentiating for extremum value we get 2A+B=pi (or 0 which gets rejected)

keeping A constant and diff for extremum value we get 2B+A=pi

this implies A+B= 2(pi)/3.....(ii)

similarly using equation (i) we can get A+C=2(pi)/3....(iii) and B+C=2(pi)/3....(iv)

using (ii),(iii) and (iv) we can see that triangle is equilateral

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