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

Answer: (B)

$\frac{4}{3}$

$\cos A+\cos B+\cos C=\frac{7}{4}$
$\Rightarrow 2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)+1-{\sin }^2\frac{C}{2}=\frac{7}{4}$
$\Rightarrow 2\cos (\frac{\pi }{2}-\frac{C}{2})\cos \left(\frac{A-B}{2}\right)-{\sin }^2\frac{C}{2}=\frac{3}{4}$
$\Rightarrow 2\sin \frac{C}{2}(\cos \left(\frac{A-B}{2}\right)-\sin (\frac{\pi }{2}-\frac{A+B}{2}))=\frac{3}{4}$
$\Rightarrow 2\sin \frac{C}{2}(\cos \left(\frac{A-B}{2}\right)-\cos \left(\frac{A+B}{2}\right))=\frac{3}{4}$
$\Rightarrow 4\sin \frac{C}{2}\sin \frac{A}{2}\sin \frac{B}{2}=\frac{3}{4}$
$\Rightarrow 4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}=\frac{3R}{4}$
$\Rightarrow r=\frac{3R}{4}$
Therefore, $\frac{R}{r}=\frac{4}{3}$
Ans: B