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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... enter image description here

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  • $\begingroup$ can you pls explain why do you need a perpendicular to SBC so badly? $\endgroup$ – dmytro Nov 20 '13 at 17:13
  • $\begingroup$ Well, after getting interested in this, I think the reason is the definition of "angle between line and plane in three dimensional space": one has to take a perpendicular from the line to the plane and then join both intersection points ( the original line's and the perpendicular's) with the plane and the angle between this segment (nothing but the line's projection on the plane) and the original line is, by definition, the wanted angle. The problem here is that the "natural" line from line $\;SO\;$ to the plane $\;SBC\;$, namely $\;SE\;$ definitely isn't perpendicular to the plane $\;SBC\;$ $\endgroup$ – DonAntonio Nov 20 '13 at 17:15
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If I understand the very question correctly, you can do the following:

  1. 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)
  2. 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.)
  3. Let OK be such perpendicular line built from O until its intersection with SE at point K
  4. 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.
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  • $\begingroup$ Excellent, ausgezeichnet, wunderbar!+1 $\endgroup$ – DonAntonio Nov 21 '13 at 5:06
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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 $SBC$.

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  • $\begingroup$ Well, no. That's the problem: $\;SE\;$ definitely isn't the (orthogonal or perpendicular, of course) projection of $\;SO\;$ on the plane, as $\;OE\;$ is not perpendicular to the plane since $\;\angle SEO\neq 90^\circ\;$, certainly...and we do need the perpendicular projection of $\;SO\;$ on the plane by definition, or at least prove this projection from $\;O\;$ intersects the line $\;SE\;$ . $\endgroup$ – DonAntonio Nov 20 '13 at 17:24
  • $\begingroup$ You're making confusion here. $SE$ is definetely the projection of $SO$ on $SBC$, because $EO$ is perpendicular to $BC$ and $O$ and $S$ are at the same distance from $B$ and $C$. The fact that the triangle is isosceles accounts for that by simmetry. $\endgroup$ – Mateus Sampaio Nov 20 '13 at 17:35
  • $\begingroup$ I just can't see it: I know that $\;OB=OC\;,\;\;SB=SC\;$, fine. How from this follows that $\;SE\;$ is the projection of $\;SO\;$ on $\;SBC\;$ ?! Or even better (and perhaps simpler): if this is true, why does the perpendicular to plane $\;SBC\;$ from $\;SO\;$ intersects the line $\;SE\;$ , which is my problem? Where do we have a contradiction if that perpendicular wouldn't meet the line $\;SE\;$ ? Thanks for your input. $\endgroup$ – DonAntonio Nov 21 '13 at 0:10

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