A Surprising Symmetry

Relevant wiki: Solving Triangles - Problem Solving - Medium

Let the required expression be $\mathfrak{P}$ . Use $\small{\color{#D61F06}{b=2R\sin B, c=2R \sin C}}$ and then $\small{\color{#3D99F6}{\sin B-\sin C=2\sin \left(\dfrac{B-C}{2}\right)\underbrace{\cos\left(\dfrac{B+C}{2}\right)}_{\large{\sin}\left(\frac{A}{2}\right)}}}$

$\mathfrak P=\dfrac{4R\left(\cancel{\sin\left(\dfrac{B-C}{2}\right)}\sin\left(\dfrac{A}{2}\right)\right)}{\cancel{\sin\left(\dfrac{B-C}{2}\right)}}$

$\boxed{\color{#007fff}{\small{\begin{aligned}\sin\left(\dfrac{A}{2}\right)=\sqrt{\dfrac{1-\color{teal}{\cos A}}{2}}=&\sqrt{\dfrac{1-\color{teal}{\sqrt{1-\color{#20A900}{\sin^2A}}}}{2}}\\&=\sqrt{\dfrac{1-\color{teal}{\sqrt{1-\color{#20A900}{\dfrac{a^2}{4R^2}}}}}{2}} \end{aligned}}}}$

Thus,

$\large\mathfrak P=4R\left( \sqrt{\dfrac{1-\color{teal}{\sqrt{1-\color{#20A900}{\dfrac{a^2}{4R^2}}}}}{2}}\right)$

Substituting $a=10, R=13$ , we get:

$\Large \mathfrak P=2\sqrt{26}=\sqrt{104}$

$\huge \therefore n=\boxed{104}$