How do I find the surface area of an angled conic base? Thank you for viewing my question.
I need help creating a formula for finding the surface area of a conic base. (eg. I install a flood light on my roof, I want to know how much surface area it will cover).
What I know:


*

*25ft - The height of the light source above the ground

*30 degrees - The angle at which the light source is pointed towards the earth

*25 degrees - The vertical light beam width

*25 degrees The horizontal light beam width 


Vertical light beam width is the angle in which the light emits horizontally from the lens of the light source. This creates a left and right limit.
Horizontal light beam width is the angle in which the light emits vertically from the lens of the light source. This creates a top and bottom limit. 
example of vertical light beam width and horizontal light beamwidth
 A: The illuminated region will be an ellipse. If the length and width of this ellipse are $a$ and $b$, then it's area is $\pi a b/4$.
If you don't know $a$ and $b$, then they will need to be calculated from the known quantities. I don't know what you mean by "vertical light beam width" and "horizontal light beam width", so I can't give any further help. A picture might help explain.
If these things are, in fact, $a$ and $b$, then we're done.
A: 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 $A \approx 1123\, \text{ft}^2$.
