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

In any $\triangle ABC$, the expression $\frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4b^2c^2}$ is equal to:

In a $\Delta ABC$,the expression $\left( \frac{a^2}{\sin A} + \frac{b^2}{\sin B} + \frac{c^2}{\sin C} \right) \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$ simplifies to,where $\Delta$ is the area of the triangle.

In $\triangle ABC$,if $(a-b)(s-c)=(b-c)(s-a)$,then $r_1, r_2, r_3$ are in

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

In a triangle $ABC$,with usual notations,if $a=4, b=8, \angle C=60^{\circ}$,then the value of $\angle B$ and the ratio $\cos A : \cos C$ respectively are,