If $a=2, b=3, c=4$ in a triangle $ABC$,then $\cos C=$

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 a $\triangle ABC$,if $\frac{a}{b^2-c^2} + \frac{c}{b^2-a^2} = 0$,then $B$ is equal to

The area of the triangle $ABC$ is $10\sqrt{3} \text{ cm}^2$,angle $B$ is $60^{\circ}$ and its perimeter is $20 \text{ cm}$. Then $\ell(AC) = $ (in $\text{ cm}$)

In a $\triangle ABC$,the expression $\frac{\cos C+\cos A}{c+a}+\frac{\cos B}{b}$ is equal to

Let $ABC$ be a triangle such that $AB = BC$. Let $F$ be the mid-point of $AB$ and $X$ be a point on $BC$ such that $FX$ is perpendicular to $AB$. If $BX = 3XC$,then the ratio $BC / AC$ equals