How many triangles have area $ 10$ and vertices at $ (\minus{}5,0)$, $ (5,0)$, and $ (5\cos \theta, 5\sin \theta)$ for some angle $ \theta$?

$ \textbf{(A)}\ 0\qquad
\textbf{(B)}\ 2\qquad
\textbf{(C)}\ 4\qquad
\textbf{(D)}\ 6\qquad
\textbf{(E)}\ 8$
1. We are given three vertices of a triangle: \( A(-5, 0) \), \( B(5, 0) \), and \( C(5\cos \theta, 5\sin \theta) \). We need to find how many such triangles have an area of 10.

2. The base of the triangle is the distance between points \( A \) and \( B \). This distance can be calculated using the distance formula:
   \[
   AB = \sqrt{(5 - (-5))^2 + (0 - 0)^2} = \sqrt{10^2} = 10
   \]

3. The area of a triangle with base \( b \) and height \( h \) is given by:
   \[
   \text{Area} = \frac{1}{2} \times b \times h
   \]
   Given that the area is 10 and the base \( AB \) is 10, we can set up the equation:
   \[
   10 = \frac{1}{2} \times 10 \times h
   \]
   Solving for \( h \):
   \[
   10 = 5h \implies h = 2
   \]

4. The height of the triangle is the perpendicular distance from point \( C \) to the line \( AB \). Since \( AB \) lies on the x-axis, the height is simply the y-coordinate of point \( C \), which is \( 5\sin \theta \).

5. We need \( 5\sin \theta = 2 \). Solving for \( \sin \theta \):
   \[
   \sin \theta = \frac{2}{5}
   \]

6. The sine function has two solutions in the interval \( [0, 2\pi) \) for each value of \( \sin \theta \). Therefore, \( \theta \) can be:
   \[
   \theta = \arcsin\left(\frac{2}{5}\right) \quad \text{or} \quad \theta = \pi - \arcsin\left(\frac{2}{5}\right)
   \]

7. Additionally, since the circle is symmetric, the same values of \( \theta \) will also occur in the intervals \( [\pi, 2\pi) \), giving us:
   \[
   \theta = \pi + \arcsin\left(\frac{2}{5}\right) \quad \text{or} \quad \theta = 2\pi - \arcsin\left(\frac{2}{5}\right)
   \]

8. Therefore, there are 4 distinct values of \( \theta \) that satisfy the condition \( 5\sin \theta = 2 \).

Conclusion:
\[
\boxed{4}
\]