Question

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

• A. $\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}=\frac{P}{2\sqrt{k}}$
• B. $a\cos A+b\cos B+c\cos C=\frac{4q-p^2}{2\sqrt{k}}$
• C. $a\sin A+b\sin B+c\sin C=\frac{2p\Delta }{\sqrt{k}}$
• D. $\sin A\sin B\sin C=\frac{8{\Delta }^3}{k}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}=\frac{P}{2\sqrt{k}}$
B $a\cos A+b\cos B+c\cos C=\frac{4q-p^2}{2\sqrt{k}}$
C $a\sin A+b\sin B+c\sin C=\frac{2p\Delta }{\sqrt{k}}$
D $\sin A\sin B\sin C=\frac{8{\Delta }^3}{k}$

$\left(a\right)\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}=\frac{b^2+c^2-a^2}{2abc}+\frac{c^2+a^2-b^2}{2abc}+\frac{a^2+b^2-c^2}{2abc}\frac{a^2+b^2+c^2}{2abc}=\frac{P}{2\sqrt{k}}\left(b\right)a\cos A+b\cos B+c\cos C=\frac{a^2(b^2+c^2-a^2)+b^2(c^2+a^2-b^2)+c^2(a^2+b^2-c^2)}{2abc}=\frac{4(a^2{b}^2+b^2{c}^2+c^2{a}^2)-(a^2+b^2+c^2{)}^2}{2abc}=\frac{4q-p^2}{2\sqrt{k}}$
$\left(c\right)a\sin A+b\sin B+c\sin C=\frac{a^2+b^2+c^2}{2R}=\frac{(a^2+b^2+c^2)4\Delta }{2abc}=\frac{2p\Delta }{\sqrt{k}}.\left(d\right)\sin A\sin B\sin C=\frac{abc}{8R^3}=\frac{abc\times 64{\Delta }^3}{8(abc{)}^3}=\frac{8{\Delta }^3}{k}.$