Question

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

• A. an isosceles
• B. an equilateral
• C. a scalene
• D. a right angled

Answer 1

The correct option is B an equilateral

In a triangle
$A+B+C=\pi$
$\cos A+\cos B+\cos C=2\cos \frac{A+B}{2}\cos \frac{A-B}{2}+\cos C=2\cos (\frac{\pi }{2}-\frac{C}{2})\cos \frac{A-B}{2}+1-2{\sin }^2\frac{C}{2}=-2{\sin }^2\frac{C}{2}+2\sin \frac{C}{2}\cos \frac{A-B}{2}+1$
$\Rightarrow -2{\sin }^2\frac{C}{2}+2\sin \frac{C}{2}\cos \frac{A-B}{2}+1=\frac{3}{2}\Rightarrow 4{\sin }^2\frac{C}{2}-4\sin \frac{C}{2}\cos \frac{A-B}{2}+1=0\cdots \left(1\right)$

Now, we know that
$\sin \frac{C}{2}\in R$
$\Delta \ge 0\Rightarrow 16{\cos }^2\frac{A-B}{2}-16\ge 0\Rightarrow {\cos }^2\frac{A-B}{2}-1\ge 0\Rightarrow {\cos }^2\frac{A-B}{2}\ge 1$
But ${\cos }^2\theta \le 1$, so
${\cos }^2\frac{A-B}{2}=1\Rightarrow A=B$
Using equation $\left(1\right)$,
$4{\sin }^2\frac{C}{2}-4\sin \frac{C}{2}+1=0\Rightarrow {(2\sin \frac{C}{2}-1)}^2=0\Rightarrow \sin \frac{C}{2}=\frac{1}{2}\Rightarrow C=\frac{\pi }{3}$
$\therefore A=B=C=\frac{\pi }{3}$

Alternate solution:
By observation,
$\cos A=\cos B=\cos C=\frac{1}{2}$
Satisfies the given equation
$\Rightarrow A=B=C={60}^{\circ }$
Therefore, the triangle is equilateral triangle.