Question

lf 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 which of the following is correct?

• 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. $\sin A\sin B\cos C=\frac{8{\Delta }^3}{k}$

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 $\sin A\sin B\cos C=\frac{8{\Delta }^3}{k}$

$x^3-p{x}^2+qx-k=0$

$a^2+b^2+c^2=p$ , $a^2{b}^2+b^2{c}^2+c^2{a}^2=q$ , $a^2{b}^2{c}^2=k$

(A) $\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}=\frac{b^2+c^2-a^2}{2abc}+\frac{c^2+a^2-b^2}{2abc}+\frac{a^2+b^2-c^2}{2abc}$

=$\frac{a^2+b^2+c^2}{2abc}$

=$\frac{p}{2\sqrt{k}}$ (correct)

(B) $a\cos A+b\cos B+c\cos C=\frac{a{b}^2+a{c}^2-a^3}{2bc}+\frac{b{c}^2+a^{2}b-b^3}{2ac}+\frac{a^{2}c+b^{2}c-c^3}{2ab}$

=$\frac{a^2{b}^2+a^2{c}^2-a^4+b^2{c}^2+a^2{b}^2-b^4+a^2{c}^2+b^2{c}^2-c^4}{2abc}$

=$\frac{4q-p^2}{2\sqrt{k}}$ (correct)

(C) $a\sin A+b\sin B+c\sin C=\frac{a^2}{2R}+\frac{b^2}{2R}+\frac{c^2}{2R}$

=$\frac{a^2+b^2+c^2}{2R}$ $(R=\frac{a}{2\sin A}=\frac{b}{2\sin B}=\frac{c}{2\sin C}=\frac{abc}{4△})$

=$\frac{(a^2+b^2+c^2)4△}{2abc}$

=$\frac{2p△}{\sqrt{k}}$

(D) $\sin A\sin B\sin C=\frac{a}{2R}\times \frac{b}{2R}\times \frac{c}{2R}=\frac{abc}{8R^3}$

=$\frac{abc}{8\times (abc{)}^3}\times 64\times (△{)}^3$

=$\frac{8{△}^3}{(abc{)}^2}\Rightarrow \frac{8{△}^3}{k}$ (correct)

Hence, $\dfrac { \cos { A } }{ a } +\dfrac { \cos { B } }{ b } +\dfrac { \cos { C } }{ c