Question

In a triangle ABC, $\frac{r_1+r_2}{1+cosC}$ is equal to

• A. $\frac{2abc}{\Delta }$
• B. $\frac{a+b}{c\Delta }$
• C. $\frac{abc}{2\Delta }$
• D. $\frac{abc}{{\Delta }^2}$

Answer 1

The correct option is C $\frac{abc}{2\Delta }$

we know that $r_1=4R\sin \frac{A}{2}\cos \frac{B}{2},r_2=\sin \frac{B}{2}\cos \frac{A}{2}\cos \frac{C}{2}$

Let, $t=\frac{r_1+r_2}{1+cosC}=\frac{4R(\sin \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}+\sin \frac{B}{2}\cos \frac{A}{2}\cos \frac{C}{2})}{2{\cos }^2\frac{C}{2}}$
$\Rightarrow t=\frac{2R(\sin \frac{A}{2}\cos \frac{B}{2}+\sin \frac{B}{2}\cos \frac{A}{2})}{\cos \frac{C}{2}}$
$\Rightarrow t=\frac{2R\sin \left(\frac{A+B}{2}\right)}{\cos \frac{C}{2}}=\frac{2R\sin \left(\frac{A+B}{2}\right)}{\cos (\frac{\pi }{2}-\frac{A+B}{2})}=2R$
$\Rightarrow t=\frac{abc}{2△}$
Ans: C