$25 \cdot 18$ For some angle $\theta$, the number of triangles with vertices at $(-5,0),(5,0)$ and $(5 \cos \theta, 5 \sin \theta)$ that have an area of 10 is
(A) 0.
(B) 2.
(C) 4.
(D) 6.
(E) 8.
(49th American High School Mathematics Examination, 1998)
［Solution］Given $|BC|=10$, and
the height $AD=|5 \sin \theta$|. (as shown in the figure).
Also, $S_{\triangle}=10=\frac{1}{2} \cdot 10 \cdot AD$,
we know $|5 \sin \theta|=2, 0<\theta<2\pi$,

Solving this, we get
$$
\theta=\sin^{-1} \frac{2}{5}, \theta=\pi \pm \sin^{-1} \frac{2}{5}, \theta=2\pi-\sin^{-1} \frac{2}{5}.
$$

There are four solutions in total (the conclusion is geometrically obvious). Therefore, the answer is $(C)$.