Question

If in $△ABC,\cos A+\cos B+\cos C=3/2$, then triangle ABC is

• A. right angled
• B. isosceles
• C. acute
• D. equilateral

Answer 1

The correct option is D equilateral

$\cos A+\cos B+\cos C=3/2$
$\Rightarrow 2\left(2\cos \frac{A+B}{2}\cos \frac{A-B}{2}\right)+2(1-2{\sin }^2\frac{C}{2})=3$
As $A+B+C=\pi$
$\Rightarrow \left(4\sin \frac{C}{2}\cos \frac{A-B}{2}\right)+(2-4{\sin }^2\frac{C}{2})=3$
$\Rightarrow 4{\sin }^2\frac{C}{2}-4\cos \left(\frac{A-B}{2}\right)\sin \frac{C}{2}+1=0$
Since, $D\ge 0$
$16{\cos }^2\left(\frac{A-B}{2}\right)-4\left(4\right)\left(1\right)\ge 0$
$\Rightarrow {\cos }^2\left(\frac{A-B}{2}\right)\ge 1$
But ${\cos }^{2}x\in [0,1]$
So, ${\cos }^2\left(\frac{A-B}{2}\right)=1$
$\Rightarrow \frac{A-B}{2}=0\Rightarrow A=B$
Similarly, $B=C$
Therefore, $A=B=C$
Hence, It is an equilateral triangle