Two sides of a triangle are $\sqrt{3}+1$ and $\sqrt{3}-1$ and the included angle is $60^{\circ}$. Find the difference of the remaining angles. (in $^{\circ}$)

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

In $\triangle ABC$,if $\cot \frac{A}{2} : \cot \frac{B}{2} : \cot \frac{C}{2} = 3 : 7 : 9$,then $a : b : c =$

In a triangle,the lengths of the two larger sides are $10 \, cm$ and $9 \, cm$ respectively. If the angles of the triangle are in $A.P.$,then the length of the third side in $cm$ can be

The area of triangle $ABC$,in which $a = 1$,$b = 2$,and $\angle C = 60^\circ$ is

In a triangle,the lengths of the sides are integers. Suppose that the length of one side is $1$,and the longest altitude is twice the shortest altitude. Let $R$ and $r$ be the circumradius and inradius of the triangle,respectively. If $R:r = m:n$,where $m$ and $n$ are coprime positive integers,then $m + n$ is