Question

In a triangle $ABC$ if $\cos A+\cos B+\cos C=\frac{3}{2},$ then the triangle is

• A. Right angled
• B. Right angled isoceles
• C. Equilateral
• D. None of these

Answer 1

The correct option is C Equilateral

$\cos A+\cos B+\cos C=\frac{3}{2}\Rightarrow 2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)+1-2{\sin }^2\frac{C}{2}=\frac{3}{2}\Rightarrow 2{\sin }^2\frac{C}{2}-2\sin \frac{C}{2}\cos \left(\frac{A-B}{2}\right)+\frac{1}{2}=0$

As $\sin \frac{C}{2}\in R$
$\Delta \ge 0\left(\Delta \text{is discriminant of quadratic}\right)$
$\Rightarrow 4{\cos }^2\left(\frac{A-B}{2}\right)-4\ge 0\Rightarrow {\cos }^2\left(\frac{A-B}{2}\right)\ge 1$
$\cos \left(\frac{A-B}{2}\right)\ge 1$ or,
$\cos \left(\frac{A-B}{2}\right)\le -1$
So the only possible solution will occur when,
$\cos \left(\frac{A-B}{2}\right)=1\Rightarrow A=B$
Therefore the solution of the quadratic equation will be,
$2{\sin }^2\frac{C}{2}-2\sin \frac{C}{2}+\frac{1}{2}=0\Rightarrow {(\sin \frac{C}{2}-\frac{1}{2})}^2=0\Rightarrow \frac{C}{2}=\frac{\pi }{6}\Rightarrow C=\frac{\pi }{3}$
Hence the triangle will be equilateral.