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

The smallest angle of the triangle whose sides are $6+\sqrt{12}, \sqrt{48}, \sqrt{24}$ is

If in a $\Delta ABC$,the altitudes from the vertices $A, B, C$ on opposite sides are in $H.P.$,then $\sin A, \sin B, \sin C$ are in

If in a triangle $ABC$,$\frac{\sin A}{4} = \frac{\sin B}{5} = \frac{\sin C}{6}$,then the value of $\cos A + \cos B + \cos C$ is equal to

Let $ABC$ be an acute-angled triangle with area $R$. Then,$\sqrt{a^2 b^2-4 R^2}+\sqrt{b^2 c^2-4 R^2}+\sqrt{c^2 a^2-4 R^2} = $

The sides of a triangle are $\sin \alpha$,$\cos \alpha$,and $\sqrt{1 + \sin \alpha \cos \alpha}$ for some $0 < \alpha < \frac{\pi}{2}$. Then the greatest angle of the triangle is.....$^o$