In any triangle $ABC$,what is the value of $a(b \cos C - c \cos B)$?

In a triangle,the lengths of the two larger sides are $10 \, cm$ and $9 \, cm$ respectively. If the angles of the triangle are in $A.P.$,then the length of the third side in $cm$ can be

Given $\frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13}$ for a $\Delta ABC$ with usual notation. If $\frac{\cos A}{\alpha} = \frac{\cos B}{\beta} = \frac{\cos C}{\gamma}$,then the ordered triple $(\alpha, \beta, \gamma)$ is proportional to

The area of triangle $ABC$,in which $a = 1$,$b = 2$,and $\angle C = 60^\circ$ is

In $\triangle ABC$,with usual notations,if $a^4+b^4+c^4-2a^2c^2-2c^2b^2=0$,then $\angle C = \dots$ (in $^{\circ}$)

In a $\triangle ABC$,$b:c = \sqrt{3}:\sqrt{2}$ and the angles $A, B, C$ are in $AP$,then $\angle A = $ (in $^{\circ}$)