Question

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

Answer 1

In a $△ABC,A+B+C={180}^0$
$\cos A+\cos B+\cos C=\frac{7}{4}$
$\Rightarrow 2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)+1-2{\sin }^2\left(\frac{C}{2}\right)=\frac{7}{4}$
$\Rightarrow 2\cos (\frac{\pi }{2}-\frac{C}{2})\cos \left(\frac{A-B}{2}\right)+1-2{\sin }^2\left(\frac{C}{2}\right)=\frac{7}{4}$
$\Rightarrow 2\sin \left(\frac{C}{2}\right)\cos \left(\frac{A-B}{2}\right)+1-2{\sin }^2\left(\frac{C}{2}\right)=\frac{7}{4}$
$\Rightarrow 2\sin \left(\frac{C}{2}\right)[\cos \left(\frac{A-B}{2}\right)-\sin \left(\frac{C}{2}\right)]=\frac{7}{4}-1$
$\Rightarrow 2\sin \left(\frac{C}{2}\right)[\cos \left(\frac{A-B}{2}\right)-\sin (\frac{\pi }{2}-\frac{A+B}{2})]=\frac{3}{4}$
$\Rightarrow \sin \left(\frac{C}{2}\right)[\cos \left(\frac{A-B}{2}\right)-\cos \left(\frac{A+B}{2}\right)]=\frac{3}{8}$
$\Rightarrow \sin \left(\frac{C}{2}\right)\times 2\sin \frac{A}{2}\sin \frac{B}{2}=\frac{3}{8}$ using transformation angle formula $\cos C-\cos D$
$\Rightarrow \sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}=\frac{3}{16}$
So that, $r=4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$
$=4\times \frac{3}{16}R$
$\Rightarrow \frac{r}{R}=4\times \frac{3}{16}=\frac{3}{4}$
Hence, $\frac{R}{r}=\frac{4}{3}$