In a $\triangle ABC$,if $A-B=120^{\circ}$ and $R=8r$,then find the value of $\frac{1+\cos C}{1-\cos C}$.

In $\triangle ABC$,if $\sin^2 B = \sin C$ and $3 \cos^2 B = 2 \cos^2 C$,then $\triangle ABC$ is

Let $p_1, p_2, p_3$ be the altitudes of a triangle $ABC$ drawn through the vertices $A, B, C$ respectively. If $r_1=4, r_2=6, r_3=12$ are the ex-radii of triangle $ABC$,then $\frac{1}{p_1^2}+\frac{1}{p_2^2}+\frac{1}{p_3^2}=$

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:

$ABCD$ is a trapezium such that $AB$ and $CD$ are parallel and $BC \perp CD$. If $\angle ADB = \theta$,$BC = p$ and $CD = q$,then $AB$ is equal to:

In a triangle $ABC$,$\angle B < \angle C$ and the values of $B$ and $C$ satisfy the equation $2 \tan x - k (1 + \tan^2 x) = 0$,where $0 < k < 1$. Then the measure of angle $A$ is: