$13 \cdot 58$ In $\triangle A B C$, if
$(\sin B+\sin C):(\sin C+\sin A):(\sin A+\sin B)=4: 5: 6$. Then the degree of the largest angle of the triangle is
(A) $72^{\circ}$.
(B) $84^{\circ}$.
(C) $1.5^{\circ}$.
(D) $120^{\circ}$.
(E) $150^{\circ}$.
(Chongqing, Sichuan Province, China Junior High School Mathematics Competition, 1984)
［Solution］By the Law of Sines and the given conditions,
$$
\begin{array}{r}
(b+c):(c+a):(a+b)=4: 5: 6 . \\
\text { Let } \quad b+c=4 k, \quad c+a=5 k, \quad a+b=6 k,
\end{array}
$$

Adding the three equations and simplifying, we get $\quad a+b+c=7.5 k$.
Thus, $a=3.5 k, b=2.5 k, c=1.5 k$.
Let $k=2 t$, then $a=7 t, b=5 t, c=3 t$, and we know the largest angle is $A$. Also,
$$
\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{\left(3^{2}+5^{2}-7^{2}\right) t^{2}}{2 \cdot 3 \cdot 5 \cdot t^{2}}=-\frac{1}{2},
$$

so $A=120^{\circ}$.
Therefore, the answer is $(D)$.