In $\Delta ABC$,$(a-b)^2 \cos^2 \frac{C}{2} + (a+b)^2 \sin^2 \frac{C}{2} =$

Let $ABC$ be an isosceles triangle with $BC$ as its base. Then,$r_1=$

In a $\Delta ABC$,if ${c^2} + {a^2} - {b^2} = ac$,then $\angle B = $

The sides of a triangle are in the ratio $1 : \sqrt{3} : 2$. Then the angles are in the ratio

Let $ABC$ be a triangle such that $AB = BC$. Let $F$ be the mid-point of $AB$ and $X$ be a point on $BC$ such that $FX$ is perpendicular to $AB$. If $BX = 3XC$,then the ratio $BC / AC$ equals

In a triangle $ABC$,if $a = 2$,$B = 60^\circ$ and $C = 75^\circ$,then $b =$