Question

If a, b, c are the sides of the triangle ABC and $a^2,b^2,c^2$ are the roots of $x^3-p{x}^2+qx-k=0$

Answer 1

As $a^2,b^2,c^2$ are roots of $x^3-p{x}^2+qx-k=0$
Gives
$a^2+b^2+c^2=p{a}^2{b}^2+b^2{c}^2+a^2{c}^2=q{a}^2{b}^2{c}^2=k$
A) $\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}=\frac{1}{2}\Rightarrow \frac{-a^2+b^2+c^2}{2abc}+\frac{a^2-b^2+c^2}{2abc}+\frac{a^2+b^2-c^2}{2abc}=\frac{1}{2}$
$\frac{p}{2\sqrt{k}}=\frac{1}{2}\Rightarrow p=k^2$
B) $a\cos A+b\cos B+c\cos C=0\Rightarrow a\left(\frac{-a^2+b^2+c^2}{2bc}\right)+b\left(\frac{a^2-b^2+c^2}{2ac}\right)+c\left(\frac{a^2+b^2-c^2}{2ab}\right)=0$
$\Rightarrow \frac{4q-p^2}{2\sqrt{k}}=0\Rightarrow 4q=p^2$
C) $\sin 2A+\sin 2B+\sin 2C=0\Rightarrow 2\sin A\cos A+2\sin B\cos B+2\sin C\cos C=0$
Substituting $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{2R}$, we get
$\frac{1}{R}(a\cos A+b\cos B+c\cos C)=0\Rightarrow \frac{4q-p^2}{2\sqrt{k}}=0\Rightarrow 4q=p^2$
D) $a\sin A+b\sin B+c\sin C=2△\Rightarrow \frac{1}{2R}(a^2+b^2+c^2)=2△\Rightarrow \frac{2p△}{\sqrt{k}}=2△\Rightarrow p=k^2$