In a triangle $ABC$,$\frac{a}{b}=2+\sqrt{3}$ and $\angle C=60^{\circ}$. Then the measure of $\angle A$ is (in $^{\circ}$)

In $\triangle ABC$, with usual notations, if the sides $a, b$ and $c$ are in the ratio $18 : 17 : 7$, then $\cot \frac{A}{2} : \cot \frac{B}{2} : \cot \frac{C}{2} = $

In a $\triangle ABC$,the angle bisector $BD$ of $\angle B$ intersects $AC$ in $D$. Suppose $BC=2$,$CD=1$ and $BD=\frac{3}{\sqrt{2}}$. The perimeter of the $\triangle ABC$ is

Let $ABC$ and $ABC^{\prime}$ be two non-congruent triangles with sides $AB=4$,$AC=AC^{\prime}=2\sqrt{2}$ and $\angle B=30^{\circ}$. The absolute value of the difference between the areas of these triangles is

In a $\Delta ABC,$ if $(\sin A + \sin B + \sin C)(\sin A + \sin B - \sin C) = 3\sin A\sin B,$ then the angle $C$ is equal to

In any $\triangle ABC$,$a(b \cos C - c \cos B)$ equals