The sum of the sides of a right-angled triangle is $42$,and the difference between the median and altitude drawn from the vertex at the right angle is $2$. The area of the triangle is

In a $\Delta ABC,$ let $\angle C = \frac{\pi}{2}.$ If $r$ and $R$ are the inradius and the circumradius respectively of the triangle,then $2(r + R)$ is equal to

In an isosceles triangle $ABC$,$\angle C = \angle A$. If the point of intersection of the bisectors of internal angles $\angle A$ and $\angle C$ divides the median of side $AC$ in the ratio $3 : 1$ (from vertex $B$ to side $AC$),then the value of $\csc \frac{B}{2}$ is equal to

In a $\triangle ABC$,if $a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = \frac{3b}{2}$,then

$A$ triangle $ABC$ has area of $P$ square units and perimeter $2S$ units. If $h_1, h_2$ and $h_3$ are respectively the lengths of the altitudes of the triangle drawn from the vertices $A, B$ and $C$,then $P^2 \left[ \frac{(h_1 h_2 + h_2 h_3 + h_3 h_1)^2}{h_1^2 h_2^2 h_3^2} - 2 \right] =$

In $\triangle ABC$,if $\cos 3A + \cos 3B + \cos 3C + \cos 3\pi = 0$,then the least value of the sum of two of its angles is