Question

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

• A. $\frac{4}{7}$
• B. $\frac{4}{3}$
• C. $\frac{3}{4}$
• D. None of the above

Answer 1

The correct option is B

$\frac{4}{3}$

Determine the value of $\frac{R}{r}$.

We know that, If $A,B,C$ are the angles of a triangle, then $\cos A+\cos B+\cos C=1+4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$

$$\begin{gathered} \therefore 1+4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}=\frac{7}{4}\left[\because \cos A+\cos B+\cos C=\frac{7}{4},\mathrm{given}\right] \\ \Rightarrow 4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}=\frac{7}{4}-1 \\ \Rightarrow 4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}=\frac{3}{4} \end{gathered}$$

Multiplying and dividing $LHS$ with $R$.

$$\begin{gathered} \Rightarrow \frac{R\times 4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}}{R}=\frac{3}{4} \\ \Rightarrow \frac{r}{R}=\frac{3}{4}\left[\because r=4\times R\times \sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\right] \\ \therefore \frac{R}{r}=\frac{4}{3} \end{gathered}$$

Hence option (B) is correct.