In a $\triangle ABC$,if $b=2, c=3$ and $\angle B=\frac{\pi}{6}$,then $a$ satisfies the equation

$\frac{a\cos A + b\cos B + c\cos C}{a + b + c} = $

In $\triangle ABC$, with usual notations, if $a = 4$, $b = 5$, and $c = 6$, then $\cos A$ and $\cos C$ are calculated using the Law of Cosines. Find the value of $\cos C$ and $\cos A$ to determine the relationship between $\angle C$ and $\angle A$.

In a triangle $ABC$,if $B = 3C$,then the values of $\sqrt{\frac{b + c}{4c}}$ and $\frac{b - c}{2c}$ are

Let $a, b$ and $c$ denote the lengths of sides $BC, CA$ and $AB$ of $\triangle ABC$. In $\triangle ABC$,$\angle BAC = 30^{\circ}$ and $\angle ABC = 60^{\circ}$. Then $a: b: c$ is

In a $\triangle ABC$,$a=1$,$b=\sqrt{3}$ and $\angle C=\pi/6$. Then the measure of the third side $c=$