The area of the triangle $ABC$ is $10\sqrt{3} \text{ cm}^2$,angle $B$ is $60^{\circ}$ and its perimeter is $20 \text{ cm}$. Then $\ell(AC) = $ (in $\text{ cm}$)

In a triangle,the sum of lengths of two sides is $x$ and the product of the lengths of the same two sides is $y$. If $x^2 - c^2 = y$,where $c$ is the length of the third side of the triangle,then the circumradius of the triangle is

In a triangle $ABC$,$m \angle A, m \angle B, m \angle C$ are in $A$.$P$. and the lengths of the two larger sides are $10$ units and $9$ units respectively. Then the length (in units) of the third side is:

If in a $\triangle ABC$,$r_1 = 2r_2 = 3r_3$,then the perimeter of the triangle is equal to

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

In $\triangle ABC$,if $A, B, C$ are in arithmetic progression,then $\sqrt{a^2-ac+c^2} \cdot \cos \left(\frac{A-C}{2}\right) =$