If $\alpha$ and $\beta$ are different values of $x$ satisfying $a \cos x + b \sin x = c,$ then $\tan \left( \frac{\alpha + \beta}{2} \right) = $

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 $\triangle ABC$ with usual notation,$\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}$ and $a=\frac{1}{\sqrt{6}}$,then the area of the triangle is

In a right-angled triangle,if the difference between the two acute angles is $60^{\circ}$,then the ratio of the length of the hypotenuse to the length of the perpendicular drawn to the hypotenuse from its opposite vertex is: (in $: 1$)

$\alpha, \beta$ are the roots of the equation $\sin^2 x + b \sin x + c = 0$. If $\alpha + \beta = \frac{\pi}{2}$,then $b^2 - 1 =$

In a triangle $ABC$,if $c^2-a^2=b(\sqrt{3}c-b)$ and $b^2-a^2=c(c-a)$,then $\angle ACB=$ (in $^{\circ}$)