In $\Delta ABC,$ if $b = 6, c = 8$ and $\angle A = 90^\circ,$ then the circumradius $R$ is:

In a triangle $ABC$,$a : b : c = 4 : 5 : 6$. The ratio of the radius of the circumcircle to that of the incircle is

$PQR$ is an isosceles triangle with $PQ=PR$. If the radius of the circumcircle of $\triangle PQR$ is equal to the length of $PQ$,then $\angle P=$ (in $^{\circ}$)

$\frac{1}{r^2}+\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}$ equals

In a $\Delta ABC$,if $\angle C = 90^{\circ}$,$r$ and $R$ are the inradius and circumradius of the $\Delta ABC$ respectively,then $2(r+R)$ is equal to

In $\triangle ABC$,if $2R + r = r_2$,then $\angle B$ is equal to