Question

If, in a triangle $△ABC,\cos A+\cos B+\cos C=\frac{7}{4}$, then $\frac{R}{r}$ is equal to

• A. $\frac{4}{3}$
• B. $\frac{3}{4}$
• C. $\frac{2}{3}$
• D. None of these

Answer 1

The correct option is A $\frac{4}{3}$

We have, $\cos A+\cos B+\cos C=\frac{7}{4}$

we know that $\cos A+\cos B+\cos C=1+4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$

$\Rightarrow 1+4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}=\frac{7}{4}$

$\Rightarrow \sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}=\frac{3}{16}$

we also know that $r=4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$

$r=4R.\frac{3}{16}=\frac{3R}{4}$

$\therefore \frac{R}{r}=\frac{4}{3}$