7.51 $\angle A$ is an interior angle of $\triangle A B C$, and $\sin A+\cos A=\frac{7}{12}$. Then $\triangle A B C$ is
(A) an obtuse triangle.
(B) an acute non-equilateral triangle.
(C) a right triangle.
(D) an equilateral triangle.
(E) the shape cannot be determined.
(China Beijing High School Mathematics Competition, 1984)
[Solution] Given $\sin A + \cos A = \frac{7}{12}$, squaring both sides and simplifying, we get:
$$
2 \sin A \cos A = -\frac{95}{144},
$$

Since $0^{\circ} < A < 180^{\circ}$ and $\sin A > 0$, $\therefore \cos A < 0$, so $A$ is an obtuse angle, and $\triangle ABC$ is an obtuse triangle. Therefore, the answer is $(A)$.