Question

If $\cos A+\cos B+\cos C=\frac{3}{2}$, then show that the triangle is equilateral.

Answer 1

Given: $\cos A+\cos B+\cos C=\frac{3}{2}$

$2[2\cos \frac{(A+B)}{2}.\cos \frac{(A-B)}{2}]+2\cos C=3$

$2[2\cos (\frac{\pi }{2}-\frac{C}{2}).\cos \left(\frac{A-B}{2}\right)]+2[1-2\sin {}^2\frac{A}{2}]=34\sin \frac{C}{2}.\cos \frac{(A-B)}{2}+2-4\sin {}^2\frac{A}{2}=3$

$4\sin {}^2\frac{A}{2}-4\sin \frac{C}{2}.\cos \left(\frac{A-B}{2}\right)+1=0$

This is a quadratic equation $\sin \frac{C}{2}$ has real roots

$\therefore$ Discriminant $\ge 0$

${[-4\cos \frac{(A-B)}{2}]}^2-4\times 4\times 4>=0{\left[\cos (A-B)\right]}^2\ge 1$

$\cos (A-B)=1$

Since, cosine of any angle can't be $>1$

$\Rightarrow A-B=0$

$\therefore A=B$

Similarly, we can prove that $B=C$

i.e, $A=B=C$

Hence, $△ABC$ is equilateral.