Question

If any $\Delta ABC$, if $B={60}^{\circ }$, then $(c-a{)}^2-b^2+ca=$.

Answer 1

$\cos B=\frac{a^2+c^2-b^2}{2ac}\Rightarrow \cos {60}^o=\frac{a^2+c^2-b^2}{2ac}\Rightarrow \frac{1}{2}=\frac{a^2+c^2-b^2}{2ac}\Rightarrow ac=a^2+c^2-b^2\Rightarrow ac=(c-a{)}^2+2ca-b^2\Rightarrow 0=(c-a{)}^2+ca-b^2$