Solid Geometry problem with Pyramid

AdirPmon

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Nov 19, 2013
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We have a straight pyramid with a square ABCD as its base and apex S. We're given the pyramid's height 8 and the angle 48 deg. between SA and SC. I've already managed to calculate the pyramid's volume (67.66 cubic meters). and now I'm asked to find the angle between the height SO (O=center of the square base = intersection point of its diagonals) and the pyramid's face SBC. I tried the triangle SOE , E=midpoint of BC, but I can't explain why this works: I know I must draw a perpendicular to plane SBC from some point on SO, yet OE definitely isn't this perpendicular. All I need is to show such perpendicular MUST intersect the line SE at some point is it possible to express the pyramid´s volume with the wanted angle?
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Here's the easiest way I can think of to calculate that angle OSE (if we imagine a segment joining S and E):

If this is a right pyramid, <OSC = 48/2 = 24.
Then, considering triangle OSC (imagining a segment joining O and C), OC = 8*tan(24) = 3.562.
Then, considering triangle OCE is a 45-45-90 triangle, OE = OC/√2 = 2.519.
OSE = arctan(OE/SO) = arctan(2.519/8) = 17.475 degrees.
 
Even simpler:
We know the height = 8 and line OE = 2.519
Lower "slant angle" = arc tan (8 / 2.519)
=72.522

Upper "slant angle" = 90 - 72.522 = 17.477
 
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