In triangle $ABC$,if $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$,then the triangle is:

In a triangle $ABC$ with usual notations,if $3a = b + c$,then $\cot \frac{B}{2} \cdot \cot \frac{C}{2} =$

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

In $\triangle ABC$,if $B+C=72^{\circ}$,then $\left(1+\frac{a}{c}+\frac{b}{c}\right) \left(1+\frac{c}{b}-\frac{a}{b}\right)$ is equal to

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

The sides of a triangle are $6+2 \sqrt{3}$,$4 \sqrt{3}$,and $\sqrt{24}$. The tangent of the smallest angle of the triangle is