How many triangles have area $10$ and vertices at $(-5,0),(5,0)$ and $(5\cos \theta, 5\sin \theta)$ for some angle $\theta$?
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ }4 \qquad \mathrm{(D) \ }6 \qquad \mathrm{(E) \ } 8$
The triangle can be seen as having the base on the $x$ axis and height $|5\sin\theta|$. The length of the base is $10$, thus the height must be $2$. The equation $|\sin\theta| = \frac 25$ has $\boxed{4}$ solutions, one in each quadrant.



Visually, the set of points of the form $(5\cos \theta, 5\sin \theta)$ is a circle centered at $(0,0)$ with radius 5. The missing vertex of the triangle must lie on this circle. At the same time, its distance from the $x$ axis must be 2. The set of all such points are precisely the lines $y=2$ and $y=-2$, and each of these lines intersects the circle in two points.