7.58 Given a triangle with sides of lengths $a, b, c$, and $(a+b+c)(a+b-c) = 3ab$. Then the angle opposite the side of length $c$ is
(A) $15^{\circ}$.
(B) $30^{\circ}$
(C) $45^{\circ}$.
(D) $60^{\circ}$.
(E) $150^{\circ}$.
(32nd American High School Mathematics Examination, 1981;
"Zu Chongzhi Cup" Junior High School Mathematics Invitational, 1988)
[Solution] Let $\theta$ be the angle opposite to the side of length $c$ in this triangle.
By the given condition $(a+b+c)(a+b-c)=3 a b$,
that is,
$$
(a+b)^{2}-c^{2}=3 a b
$$

or
$$
a^{2}+b^{2}-a b=c^{2} \text {. }
$$

Also, by the cosine rule, we have $a^{2}+b^{2}-2 a b \cos \theta=\dot{c}^{2}$, so
$$
a b=2 a b \cos \theta \text {. }
$$

Thus,
$$
\cos \theta=\frac{1}{2} \text {, hence } \theta=60^{\circ} \text {. }
$$

Therefore, the answer is $(D)$.