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 $^{\circ}$)

Let a triangle $ABC$ be inscribed in a circle of radius $2$ units. If the $3$ bisectors of the angles $A, B$ and $C$ are extended to cut the circle at $A_1, B_1$ and $C_1$ respectively,then the value of $\left[\frac{AA_1 \cos \frac{A}{2} + BB_1 \cos \frac{B}{2} + CC_1 \cos \frac{C}{2}}{\sin A + \sin B + \sin C}\right]^2$ is:

If $p_1, p_2, p_3$ are the altitudes of a triangle $ABC$ from the vertices $A, B, C$ respectively,then with the usual notation,$\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}+\frac{1}{r^2}=$

In $\triangle ABC$,if $\tan A + \tan B + \tan C = 6$ and $\tan A \cdot \tan B = 2$,then $\tan C = \dots$

Let $S=\{x \in(-\pi, \pi) \mid x \neq 0, \pm \frac{\pi}{2}\}$. The sum of all distinct solutions of the equation $\sqrt{3} \sec x+\operatorname{cosec} x+2(\tan x-\cot x)=0$ in the set $S$ is equal to

In a triangle $ABC$,$\angle B = \frac{\pi}{3}$ and $\angle C = \frac{\pi}{4}$,and $D$ divides $BC$ internally in the ratio $1 : 3$. Then $\frac{\sin \angle BAD}{\sin \angle CAD}$ is equal to