Given $\triangle A B C$ has a perimeter of 20, area of $01 \sqrt{3}$, and $\angle A=60^{\circ}$. Find $\sin A: \sin B: \sin C$.
Seven, Hint: By the cosine theorem, we get $(b+c)^{2}-a^{2}$ $=120$. Then, we get the system of equations $\left\{\begin{aligned} b+c & =13, \\ b c & =40 \text {. Thus, we have }\end{aligned}\right.$ $\sin A: \sin B: \sin C=7: 8: 5$