In a triangle ABC , with usual notations, $(\mathrm{a}+\mathrm{b}+\mathrm{c})(\mathrm{a}+\mathrm{b}-\mathrm{c})=3 \mathrm{ab}$, then $\angle \mathrm{C}=$

With usual notations in $\triangle \mathrm{ABC}$, if $\angle \mathrm{B}=\frac{\pi}{2}$, and $\tan \frac{\mathrm{A}}{2}, \tan \frac{\mathrm{C}}{2}$ are roots of equation $\mathrm{p} x^2+\mathrm{qx}+\mathrm{r}=0$, $\mathrm{p} \neq 0$, then

The circumradius of a triangle whose sides are 10 units, 8 units and 6 units is

In a triangle ABC , with usual notations. $\frac{2 \cos \mathrm{~A}}{\mathrm{a}}+\frac{\cos \mathrm{B}}{\mathrm{b}}+\frac{2 \cos \mathrm{C}}{\mathrm{c}}=\frac{\mathrm{a}}{\mathrm{bc}}+\frac{\mathrm{b}}{\mathrm{ca}}$. Then $\angle \mathrm{A}=$