With usual notations in $\Delta ABC$,$a=3$,$c=2$ and $\sin C=\frac{2}{3}$,then $\angle A=$

The ratio of the sides of triangle $ABC$ is $1:\sqrt{3}:2$. The ratio of angles $A:B:C$ is

In triangle $ABC$, $a=2$, $b=3$ and $\sin A=\frac{2}{3}$, then $B$ is equal to (in $^{\circ}$)

In $\Delta ABC$,$c\cos (A - \alpha ) + a\cos (C + \alpha ) = $

In a triangle $ABC$,with usual notation,$\angle B = \pi/3$ and $\angle C = \pi/4$. If $D$ divides $BC$ internally in the ratio $1:3$,find the value of $\frac{\sin \angle BAD}{\sin \angle CAD}$.

In $\triangle ABC$,if $\frac{1}{b+c} + \frac{1}{c+a} = \frac{3}{a+b+c}$,then $\angle C$ is equal to: (in $^{\circ}$)