Question

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

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

Answer 1

The correct option is A an equilateral triangle

${\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C=\cos A\cos B+\cos B\cos C+\cos C\cos A$

Multiplying both sides by $2$ in the above equation, we get

$\Rightarrow 2{\cos }^{2}A+2{\cos }^{2}B+2{\cos }^{2}C=2\cos A\cos B+2\cos B\cos C+2\cos C\cos A$

$\Rightarrow 2{\cos }^{2}A+2{\cos }^{2}B+2{\cos }^{2}C-2\cos A\cos B-2\cos B\cos C-2\cos C\cos A=0$

$\Rightarrow (\cos A-\cos B{)}^2+(\cos B-\cos C{)}^2+(\cos C-\cos A{)}^2=0$

$\therefore \cos A=\cos B,\cos B=\cos C,\cos C=\cos A$

$\therefore A=B,B=C,C=A$

$A=B=C.$

$⟹\Delta ABC$ is an equilateral triangle.

Hence, option A.