17・2 In $\triangle A B C$, the sides $a, b, c$ satisfy $(a+b+c)(a+b-c)=a b$, then the angle opposite to side $A B$ is
(A) $30^{\circ}$.
(B) $60^{\circ}$.
(C) $120^{\circ}$.
(D) $150^{\circ}$.
(China Shanxi Province Taiyuan City Junior High School Mathematics Competition, 1993)
[Solution] From the given $(a+b+c)(a+b-c)=ab$, we have $a^{2}+b^{2}-c^{2}=-ab, \quad$ and $\quad \frac{a^{2}+b^{2}-c^{2}}{2ab}=-\frac{1}{2}$. By the cosine rule, $\cos \angle C=-\frac{1}{2}, \therefore \angle C=120^{\circ}$. Therefore, the answer is $(C)$.