With usual notations,the perimeter of a triangle $ABC$ is $6$ times the arithmetic mean of the sines of its angles. If $a=1$,then $\angle A=$

Points $D$ and $E$ are taken on the side $BC$ of a triangle $ABC$ such that $BD = DE = EC$. If $\angle BAD = x$,$\angle DAE = y$,and $\angle EAC = z$,then the value of $\frac{\sin(x + y)\sin(y + z)}{\sin x \sin z}$ is:

In $\triangle ABC$, with usual notations, $(a + b + c)(b + c - a)(c + a - b)(a + b - c) = 3b^2c^2$, then $\angle A = $

In any $\triangle ABC$,$a(b \cos C - c \cos B)$ equals

In $\triangle ABC$,the sum of the lengths of two sides is $x$ and the product of those lengths is $y$. If $c$ is the length of the third side and $x^2 - c^2 = y$,then the circumradius of the triangle is

Prove that in any triangle $ABC$,$\cos A = \frac{b^{2} + c^{2} - a^{2}}{2bc}$,where $a, b, c$ are the lengths of the sides opposite to the vertices $A, B, C$ respectively.