In $\triangle ABC$,$a=6 \text{ cm}$,$b=10 \text{ cm}$ and $c=14 \text{ cm}$. Then,the sum of the acute angles of the triangle is (in $^{\circ}$)

Let $ABC$ and $ABC^{\prime}$ be two non-congruent triangles with sides $AB=4$,$AC=AC^{\prime}=2\sqrt{2}$ and $\angle B=30^{\circ}$. The absolute value of the difference between the areas of these triangles is

In any $\triangle ABC$, the expression $\frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4b^2c^2}$ is equal to:

In $\Delta ABC,\, \left( {\cot \frac{A}{2} + \cot \frac{B}{2}} \right)\,\left( {a{{\sin }^2}\frac{B}{2} + b{{\sin }^2}\frac{A}{2}} \right) =$

If in $\triangle ABC$,with usual notations,$a \cdot \cos^2 \frac{C}{2} + c \cdot \cos^2 \frac{A}{2} = \frac{3b}{2}$,then

In $\triangle ABC$,if $a = 2b$ and $|A - B| = \frac{\pi}{3}$,then $\angle C = $