In a triangle $ABC$ with usual notations,if $\tan \left(\frac{B-C}{2}\right) = x \cot \frac{A}{2}$,then $x =$

The sides of a $\triangle ABC$ are given by $a=3, b=5$ and $c=3$. Then,$\cos A=$

The angles $A, B$ and $C$ of a triangle $ABC$ are in $AP$. If $b: c = \sqrt{3}: \sqrt{2}$,then the angle $A$ is (in $^{\circ}$)

Let $a, b$ and $c$ be the lengths of the sides of a triangle with its opposite angles $A, B$ and $C$ respectively. If $\angle C=60^{\circ}$,then the value of $\frac{c(a+b)+(a^2+b^2)}{(b+c)(c+a)}$ is

With the usual notation,in $\Delta ABC$,if $\angle A + \angle B = 120^{\circ}$ and $a : b = (\sqrt{3} + 1) : (\sqrt{3} - 1)$,then the ratio $\angle A : \angle B$ is

If $\triangle ABC$ is such that $\angle A=90^{\circ}$ and $\angle B \neq \angle C$,then $\frac{b^2+c^2}{b^2-c^2} \sin (B-C)$ is equal to