Question

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

• A. Equilateral triangle
• B. Right angled triangle
• C. Acute angled triangle
• D. Obtuse angled triangle

Answer 1

The correct option is A Equilateral triangle

${\cos }^{3}A+{\cos }^{3}B+{\cos }^{3}C-3\cos A\cos B\cos C=0$

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

$-\cos A\cos b-\cos B\cos C-\cos C\cos A)=0$

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

$-\cos C\cos A=0.$

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

$\Rightarrow \cos A-\cos B=0$ - and

$\cos B-\cos C=0$ and

$\cos C-\cos A=0$

$\Rightarrow \cos A=\cos B=\cos C.$

$\Rightarrow A=B=C$

$\therefore △ABC$ is equilateral triangle. opion $A$ is correct.