If two sides of a triangle are $\sqrt{3}-2$ and $\sqrt{3}+2$ units and their included angle is $60^{\circ}$,then the third side of the triangle is

With usual notation,in triangle $ABC$,$m \angle A = 30^{\circ}$. Then the value of $\left(1+\frac{a}{c}+\frac{b}{c}\right)\left(1+\frac{c}{b}-\frac{a}{b}\right)$ is equal to

In a triangle $ABC$,if $A=30^{\circ}$ and $\frac{b}{(\sqrt{3}+1)^2+2(\sqrt{2}-1)}=\frac{c}{(\sqrt{3}+1)^2-2(\sqrt{2}-1)}$,then $B=$ (in $^{\circ}$)

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.

With usual notations in $\triangle ABC$,if $\frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)}$,then $a^2, b^2, c^2$ are in

In $\triangle ABC$,$m \angle B = \frac{\pi}{3}$ and $m \angle C = \frac{\pi}{4}$. Let point $D$ divide $BC$ internally in the ratio $1:3$. Then,the value of $\frac{\sin(\angle BAD)}{\sin(\angle CAD)}$ is