Question

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

• A. isosceles
• B. right-angled
• C. equilateral
• D. none of these.

Answer 1

The correct option is C equilateral

$\cos A+\cos B+\cos C=\frac{3}{2}$
Using transformation angle formula, we have
$(\cos A+\cos B)+\cos C=\frac{3}{2}$
$\Rightarrow 2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)+\cos C=\frac{3}{2}$
Using sub-multiple angle formula to $\cos C=1-2{\sin }^2\frac{C}{2}$ we get
$\Rightarrow 2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)+1-2{\sin }^2\frac{C}{2}=\frac{3}{2}$
Since $A+B+C=\pi \Rightarrow \frac{A+B}{2}=\frac{\pi }{2}-\frac{C}{2}$
$\Rightarrow 2\cos (\frac{\pi }{2}-\frac{C}{2})\cos \left(\frac{A-B}{2}\right)-2{\sin }^2\frac{C}{2}=\frac{3}{2}-1$
$\Rightarrow 2\sin \left(\frac{C}{2}\right)\cos \left(\frac{A-B}{2}\right)-2{\sin }^2\frac{C}{2}=\frac{1}{2}$
$\Rightarrow 2\sin \left(\frac{C}{2}\right)[\cos \left(\frac{A-B}{2}\right)+\sin \left(\frac{C}{2}\right)]=\frac{1}{2}$
Again
$\Rightarrow 2\sin \left(\frac{C}{2}\right)[\cos \left(\frac{A-B}{2}\right)+\sin (\frac{\pi }{2}-\frac{A+B}{2})]=\frac{1}{2}$
$\Rightarrow 2\sin \left(\frac{C}{2}\right)[\cos \left(\frac{A-B}{2}\right)+\cos \left(\frac{A+B}{2}\right)]=\frac{1}{2}$
Using transformation angle formula, we get
$\Rightarrow \sin \left(\frac{C}{2}\right)\left[2\sin \left(\frac{B}{2}\right)\sin \left(\frac{A}{2}\right)\right]=\frac{1}{4}$
$\Rightarrow \sin \left(\frac{A}{2}\right)\sin \left(\frac{B}{2}\right)\sin \left(\frac{C}{2}\right)=\frac{1}{8}$
$\therefore \sin \left(\frac{A}{2}\right)=\frac{1}{2},\sin \left(\frac{B}{2}\right)=\frac{1}{2},\sin \left(\frac{C}{2}\right)=\frac{1}{2}$
$\Rightarrow \sin \left(\frac{A}{2}\right)=\sin \frac{\pi }{6},\sin \left(\frac{B}{2}\right)=\sin \frac{\pi }{6},\sin \left(\frac{C}{2}\right)=\sin \frac{\pi }{6}$
$\therefore \left(\frac{A}{2}\right)=\left(\frac{B}{2}\right)=\left(\frac{C}{2}\right)=\frac{\pi }{6}$
$\Rightarrow \angle A=\angle B=\angle C=2\times \frac{\pi }{6}=\frac{\pi }{3}$
Hence, the triangle is equilateral.