Question

If ${\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C=1$, then $△ABC$ is

• A. Equilateral
• B. Isosceles
• C. Scalene
• D. Right angled

Answer 1

Answer: (D)

Right angled

$A+B+C=\pi \Rightarrow B+C=\pi -A$

Given, ${\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C=1$

$\Rightarrow {\cos }^{2}A+{\cos }^{2}B=1-{\cos }^{2}C={\sin }^{2}C$

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

$\Rightarrow \cos A(\cos A-\cos (B-C\left)\right)=0$
$\Rightarrow \cos A\left(\cos \right(\pi -(B+C)-\cos (B-C))=0$
$\Rightarrow \cos A(-\cos (B+C)-\cos (B-C\left)\right)=0$
$\Rightarrow \cos A\left(2\cos B\cos C\right)=0$

$\therefore$ either $\cos A={90}^o$ or $\cos B={90}^o$ or $\cos C={90}^o$
So, its a triangle with one angle $={90}^o$
Hence, right-angled triangle