Question

In a $△ABC,{\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C=1$ prove that the triangle is right angled

Answer 1

It is given that ${\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C=1$, then,

${\cos }^{2}A+\left(1-{\sin }^{2}B\right)+{\cos }^{2}C=1$

$\left({\cos }^{2}A-{\sin }^{2}B\right)+1+{\cos }^{2}C=1$

$1+\cos \left(A+B\right)\cos \left(A-B\right)+{\cos }^{2}C=1$

$1+\cos \left(180-C\right)\cos \left(A-B\right)+{\cos }^{2}C=1$

$1-\cos C\cos \left(A-B\right)+{\cos }^{2}C=1$

$-\cos C\left(\cos \left(A-B\right)-\cos C\right)=0$

$\cos C\left(\cos \left(A-B\right)-\cos \left(180-\left(A+B\right)\right)\right)=0$

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

$\cos C\left(2\cos A\cos B\right)=0$

$\cos A\cos B\cos C=0$

$\cos A=0or\cos B=0or\cos C=0$

This gives that,

$A={90}^{\circ }orB={90}^{\circ }orC={90}^{\circ }$

Therefore, the triangle is right angled triangle.