With usual notations in $\triangle ABC$,if $a^2+b^2-c^2=ab$,then the measure of angle $C$ is

In a $\triangle ABC$,if $\angle A = 60^{\circ}$,then $(a+b+c)(b+c-a) =$

$p_1, p_2, p_3$ 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 in $\triangle ABC$,$B=45^{\circ}$,$a=2(\sqrt{3}+1)$ and the area of $\triangle ABC$ is $6+2\sqrt{3}$ sq. units,then the side $b=$

Let $ABC$ be a triangle such that $AB=4, BC=5$ and $CA=6$. Choose points $D, E, F$ on $AB, BC, CA$ respectively,such that $AD=2, BE=2, CF=2$. Then find the ratio of the area of $\triangle DEF$ to the area of $\triangle ABC$.

In a triangle $ABC$,if $\frac{\sin A - \sin C}{\cos C - \cos A} = \cot B$,then $A, B, C$ are in