Mathematics - Properties of Triangle Question with Solution | TestHub

$SI=r\Rightarrow R^2-2Rr-r^2=0$

$${\left(\frac{R}{r}\right)}^2-2\left(\frac{R}{r}\right)-1=0$$

$$\begin{array}{rl}\Rightarrow \frac{R}{r}& =\frac{2\pm \sqrt{4-4(1)(-1)}}{2}=\frac{2\pm \sqrt{8}}{2}\\ \therefore \frac{R}{r}& =\sqrt{2}+1\\ & +\frac{1}{\sqrt{2}+1}=1+\frac{\sqrt{2}-1}{(\sqrt{2}+1)(\sqrt{2}-1)}=1+\frac{\sqrt{2}-1}{2-1}=\\ & \end{array}$$

$$\therefore \cos A+\cos B+\cos C=1+\frac{r}{R}=1+\sqrt{2}-1=\sqrt{2}$$

$$\therefore [4(\cos A+\cos B+\cos C)]=[4\sqrt{2}]=5$$