Three-Dimensional geometry + trigonometry question Dear all: I read this question yesterday and it is driving me crazy! I shall offer a bounty to whoever gives a reasonable answer...
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.
Perhaps it is possible to express the pyramid´s volume by means of the wanted angle? That way we could get the angle...
 
 A: If I understand the very question correctly, you can do the following:


*

*show that the planes SOE and SBC are perpendicular (because SBC contains BC, which is normal to SOE ($BC \perp OE$ and $BC \perp SE$ which are both in SOE), and therefore, by definition, the planes SOE and SBC are perpendicular)

*now, for each point in SOE plane, you can make a unique perpendicular line to SE, which will also be perpendicular to SBC (because 1.)

*Let OK be such perpendicular line built from O until its intersection with SE at point K 

*Because SOE and SOK are in the same plane, and share the angle in question, one can say that computing $\angle OSK$ (which is a definition you're referring to) and $\angle OSE$ are equivalent.

A: First we note that as the pyramid is straight, $SO=h$ is perpendicular to the base. So the triangle $ASO$ is right, and we have $\tan \alpha=\frac{h}{AO}$, where $\alpha=48^\circ$. So, the the segment $OE$ as the half of the side is$$OE=\frac{AO}{\sqrt2}=\frac{h}{\sqrt2 \tan \alpha}.$$
Then, we have that
$$\tan \beta=\frac{OE}{SO}=\frac{h}{h\sqrt2 \tan \alpha}\Rightarrow\beta=\tan^{-1} \left(\frac{1}{\sqrt2 \tan \alpha}\right)\approx58.7^\circ,$$
where $\beta$ is the desired angle. You don't need this perpendicular you mentioned because the triangles in the pyramid are isosceles, and so $SE$ is the projection of $SO$ on the plane $SBC$.
