OK, let $\theta$ be the angle the light is tilted, $\alpha$ be the horizontal beam divergence (width), $\beta$ be the vertical beam divergence, and $h$ be the height above the ground.
You should be able to convince yourself with right triangles that the horizontal beam width on the ground is
$$h \left[\tan{\left(\theta+\frac{\alpha}{2} \right)} - \tan{\left(\theta-\frac{\alpha}{2} \right)}\right] = \frac{2 h \tan{\frac{\alpha}{2}}}{\cos^2{\theta}-\sin^2{\theta} \tan^2{\frac{\alpha}{2}}} $$
There is an analogous expression for $\beta$. The area of the beam (assuming a rectangular shape, add whatever geometrical factors you feel exist) is then
$$A = \frac{4 h^2 \tan{\frac{\alpha}{2}} \tan{\frac{\beta}{2}}}{\left(\cos^2{\theta}-\sin^2{\theta} \tan^2{\frac{\alpha}{2}}\right)\left(\cos^2{\theta}-\sin^2{\theta} \tan^2{\frac{\beta}{2}}\right)} $$
Using the given numbers, I get $A \approx 1123\, \text{ft}^2$.