Question

In triangle $ABC$, if $\cos A+\cos B+\cos C=\frac{7}{4}$, then $\frac{R}{r}$ is equal to

• A. $\frac{3}{4}$
• B. $\frac{4}{3}$
• C. $\frac{2}{3}$
• D. $\frac{3}{2}$

Answer 1

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

$\cos A+\cos B+\cos C$

$=1+4\sin \frac{A}{2}.\sin \left(\frac{B}{2}\right).\sin \frac{C}{2}$

$=\frac{7}{4}$

Or

$4\sin \frac{A}{2}.\sin \left(\frac{B}{2}\right).\sin \frac{C}{2}=\frac{3}{4}$

Now

$4\sin \frac{A}{2}.\sin \left(\frac{B}{2}\right).\sin \frac{C}{2}=\frac{r}{R}$

Hence

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