Question

If A, B, C are the angles of triangle show that system of equations
$-x+y\cos C+z\cos B=0$
$x\cos C-y+z\cos A=0$
and $x\cos B+y\cos A-z=0$
has non - zero solution.

Answer 1

For non-trivial solution we must have $△=0$

or $△=\left|\begin{array}{ccc}-1& cosC& cosB\\ cosC& -1& cosA\\ cosB& cosA& -1\end{array}\right|=0$

When $A+B+C=\pi$

i.e. $cos[A+B]=-cosC,sin(A+B)=sinC$

$△=-1(1-co{s}^{2}A)-cosC(-cosC-cosAcosB+cosB(cosAcosC+cosB\left)\right)$

= $-si{n}^{2}A+cosCsinAsinB+cosBsinAsinC$

= $-si{n}^{2}A+sinA.sin(B+C)$

= $-si{n}^{2}A+sinA.sinA=0$