Question

If in a $△ABC,\cos A+\cos B+\cos C=k+\frac{r}{R};$ where $r=$ inradius and $R=$ circumradius of $△ABC$, then find the value of $\sqrt[k+1]{k+1196}$.

Answer 1

$\cos A+\cos B+\cos C=k+\frac{r}{R}$ ...(1)

Now, $\cos A+\cos B+\cos C=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)+\cos C$

$=\sin \frac{C}{2}\cos \left(\frac{A-B}{2}\right)+1-2{\sin }^2\frac{C}{2}$

$=1+2\sin \frac{C}{2}[\cos \left(\frac{A-B}{2}\right)-\cos \left(\frac{A+B}{2}\right)]$

$=1+4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}=1+\frac{r}{R};(r=4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2})$

$\therefore$ From (1), we have $k=1$

$\therefore \sqrt[k+1]{k+1196}=\sqrt[2]{2196}=14$