Question

If in $△ABC$ ; $\frac{1+cosA}{a}+\frac{1+cosB}{b}+\frac{1+cosC}{c}=k^2\frac{(1+cosA)(1+cosB)(1+cosC)}{abc}$ then $k$ is

• A. $\frac{1}{2\sqrt{2}R}$
• B. $2R$
• C. $\frac{1}{R}$
• D. $\frac{2}{R}$

Answer 1

The correct option is B $2R$

Given,

L.H.S.$=\frac{1+\cos A}{a}+\frac{1+\cos B}{b}+\frac{1+\cos C}{c}$

$=\frac{1+\frac{b^2+c^2-a^2}{2bc}}{a}+\frac{1+\frac{c^2+a^2-b^2}{2c}}{b}+\frac{1+\frac{a^2+b^2-c^2}{2ab}}{c}$

$=\frac{b^2+c^2-a^2+2bc}{2abc}+\frac{c^2+a^2-b^2+2ac}{2abc}+\frac{a^2+b^2-c^2+2ab}{2abc}$

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

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

$=\frac{4s^2}{2abc}$

R.H.S$=k^2\frac{(1+\cos A)(1+\cos B)(1+\cos C)}{abc}$

$=k^2\times \frac{2{\cos }^2\frac{A}{2}\times 2{\cos }^2\frac{B}{2}\times 2{\cos }^2\frac{C}{2}}{abc}$

$=\frac{8k^2}{abc}\times \frac{s(s-a)}{bc}\times \frac{s(s-b)}{ac}\times \frac{s(s-c)}{ab}$

$=\frac{8k^2}{abc\times a^2{b}^2{c}^2}\times {\Delta }^2\times s^2$

L.H.S$=$R.H.S

$\therefore \frac{4s^2}{2abc}=\frac{8k^2}{abc\times a^2{b}^2{c}^2}\times {\Delta }^2\times s^2$

$\therefore k=\frac{abc}{2\Delta }$

$\therefore k=2R$.