In a triangle $ABC$,$l(AB)=\sqrt{23}$ units,$l(BC)=3$ units,$l(CA)=4$ units,then $\frac{\cot A+\cot C}{\cot B}$ is

In $\triangle ABC$,if $a^2 \sin^2 \frac{C}{2} + c^2 \sin^2 \frac{A}{2} = \frac{b^2}{2}$,then $a+c : b =$

The area of triangle $ABC$,in which $a = 1$,$b = 2$,and $\angle C = 60^\circ$ is

The angles of $\triangle ABC$ are in an arithmetic progression. If the larger sides $a, b$ satisfy the relation $\frac{\sqrt{3}}{2} < \frac{b}{a} < 1$,then the possible values of the smallest side $c$ are

In a $\triangle ABC$,$\angle C = 60^{\circ}$ and $\angle A = 75^{\circ}$. If $D$ is a point on $AC$ such that the area of $\triangle BAD$ is $\sqrt{3}$ times the area of $\triangle BCD$,then the measure of $\angle ABD$ is (in $^{\circ}$)

In $\Delta ABC$,$({b^2} - {c^2})\cot A + ({c^2} - {a^2})\cot B + ({a^2} - {b^2})\cot C = $