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

In a triangle $$\mathrm{ABC}$$, with usual notations $$\mathrm{a}=2, \mathrm{~b}=3, \mathrm{c}=5$$, then $$\frac{\cos \mathrm{A}}{\mathrm{a}}+\frac{\cos \mathrm{B}}{\mathrm{b}}+\frac{\cos \mathrm{C}}{\mathrm{c}}=$$

In $$\Delta ABC$$, with usual notations $$\mathrm{\frac{b\sin B-c\sin C}{\sin(B-C)}}=$$

With usual notations, in any $$\triangle A B C$$, if $$a\cos B=b \cos A$$, then the triangle is