In a triangle $ABC$,$a = 4$,$b = 3$,and $\angle A = 60^\circ$. Then $c$ is the root of the equation:

The lengths of the sides of a triangle are three consecutive natural numbers and its largest angle is twice the smallest one. Then the lengths of the sides of the triangle (in units) are

We are given $b, c$ and $\sin B$ such that $B$ is acute and $b < c \sin B$. Then

In a $\triangle ABC$,$2ac \sin \frac{1}{2}(A-B+C)$ is equal to

In $\triangle ABC$,if $\frac{1}{b+c} + \frac{1}{c+a} = \frac{3}{a+b+c}$,then $\angle C$ is equal to: (in $^{\circ}$)

If $\cos^2 A + \cos^2 C = \sin^2 B$,then $\Delta ABC$ is