How Do You Calculate Electric Flux Through a Pyramid's Slanted Surfaces?

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The total electric flux through the slanted surfaces of a pyramid with a square base measuring 6.00m per side and a height of 4.00m in a vertical electric field of 52.0 N/C can be calculated using the formula Φ = EAcos(θ). The angle θ is the angle between the normal to the surface and the vertical electric field. The flux through the base is equal to the flux through the slanted sides, which is confirmed by the uniform nature of the electric field across the surfaces.

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Question: A pyramid with horizontal square base, 6.00m on each side and a height of 4.00m is placed in a vertical electic field of 52.0 N/C. Calculate the total electric fluc through the pyramids four slanted surfaces.

I know you can figure out the flux through the base = 36 m * 52.0 N/C because it's a unform electric field so flux in base = flux out pyramid sides, but if you were to do it without using that fact, actually using the geometry of the sides and such, what angle should you be using and cosine or sine? I can't seem to work it out. Thanks
Ben
 
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The electric field and the angle it makes with the normal to the surface is still constant over the entire pyramid side, so the flux is \int\vec E\cdot d\vec A=EA\cos(\theta). Where theta is the angle between the normal and the vertical electric field.
 

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