In $\Delta ABC, a(b\cos C - c\cos B) = $

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} = $

$p_1, p_2, p_3$ are altitudes of a triangle $ABC$ drawn from the vertices $A, B, C$ respectively. If $\Delta$ is the area of the triangle and $2s$ is the sum of the sides,then $\frac{1}{p_1} + \frac{1}{p_2} - \frac{1}{p_3} =$

If the area of triangle $ABC$ is $b^2-(c-a)^2$,then $\tan B=$

The perimeter of $\Delta ABC$ is $6$ times the $A.M.$ of the sines of its angles. If $a = 1$,then $\angle A = $

With usual notations,if in $\triangle ABC$,$s$ is the semi-perimeter and $(s-a)(s-b)=s(s-c)$,then $\triangle ABC$ is