$7 \cdot 59$ In $\triangle A B C$, if $(a+b+c)(a+b-c)=3 a b$, and $\sin ^{2} A$ $=\sin ^{2} B+\sin ^{2} C$, then the angles of $\triangle A B C$ are
(A) $40^{\circ}, 60^{\circ}, 80^{\circ}$.
(B) $60^{\circ}, 60^{\circ}, 60^{\circ}$.
(C) $30^{\circ}, 60^{\circ}, 90^{\circ}$.
(D) $45^{\circ}, 45^{\circ}, 90^{\circ}$.
(China Shanxi Province Taiyuan City Junior High School Mathematics Competition, 1990)
[Solution 1] From $\sin ^{2} A=\sin ^{2} B+\sin ^{2} C$ and
$$
\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{2 R},
$$

thus $\sin A=\frac{a}{2 R}, \quad \sin B=\frac{b}{2 R}, \quad \sin C=\frac{c}{2 R}$,
hence $a^{2}=b^{2}+c^{2}$.
Also, from the given $(a+b)^{2}-c^{2}=3 a b$, substituting the above equation and simplifying yields
$$
2 b^{2}=a b, \quad \text { i.e., } \quad b=\frac{1}{2} a .
$$

Therefore, this triangle is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.
Thus, the answer is $(C)$.
[Solution 2] From the given $(a+b)^{2}-c^{2}=3 a b$,
we have $a^{2}+b^{2}-c^{2}=a b$.
$$
\therefore \cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}=\frac{1}{2},
$$

thus $C=60^{\circ}$, so $A+B=120^{\circ}$.
Also, $\sin ^{2} A=\sin ^{2} B+\sin ^{2} C=\sin ^{2} B+\left(1-\cos ^{2} C\right)=\sin ^{2} B+\frac{3}{4}$,

which means
$$
(\sin A+\sin B)(\sin A-\sin B)=\frac{3}{4} \text {, }
$$

or $2 \sin \frac{A+B}{2} \cos \frac{A-B}{2} \cdot 2 \sin \frac{A-B}{2} \cos \frac{A+B}{2}=\frac{3}{4}$,
$$
\therefore \quad 2 \cdot \frac{\sqrt{3}}{2} \cdot \frac{1}{2} \cdot 2 \sin \frac{A-B}{2} \cos \frac{A-B}{2}=\frac{3}{4} \text {, }
$$

thus
$$
\sin (A-B)=\frac{\sqrt{3}}{2} \text {. }
$$
$\therefore A-B=60^{\circ}$, and $A+B=120^{\circ}$, so $A=90^{\circ}, B=30^{\circ}$. Thus, the answer is $(C)$.