In a triangle $ABC$,if $\sin \frac{A}{2} = \frac{1}{4} \sqrt{\frac{3}{5}}$,$a = 2$,$c = 5$ and $b$ is an integer,then the area (in sq. units) of triangle $ABC$ is

In $\triangle ABC$,right-angled at $A$,the circumradius,inradius,and radius of the excircle opposite to $A$ are respectively in the ratio $2:5:\lambda$. Then the roots of the equation $x^2-(\lambda-5)x+(\lambda-6)=0$ are:

In a $\triangle ABC$,$\cos \left(\frac{B+2C+3A}{2}\right) + \cos \left(\frac{A-B}{2}\right)$ is equal to

If $\alpha, \beta, \gamma$ are angles of a triangle,then $\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma - 2 \cos \alpha \cos \beta \cos \gamma$ is

Consider an obtuse-angled triangle $ABC$ in which the difference between the largest and the smallest angle is $\frac{\pi}{2}$ and whose sides are in arithmetic progression. Suppose that the vertices of this triangle lie on a circle of radius $1$.
$(1)$ Let $a$ be the area of the triangle $ABC$. Then the value of $(64 a)^2$ is
$(2)$ The inradius of the triangle $ABC$ is

In $\triangle ABC$,if $\cos ^2 A + \cos ^2 B + \cos ^2 C = 1$,then $\triangle ABC$ is