Mitres on Pyramids
Date: 09/26/2002 at 09:57:07 From: Simon K Subject: Pyramid internal angles I am making a weather proofing construction for my home, and have to mitre the boards in a pyramid with a rectangular - not square - base, and an apex that is off to one side (though central in relation to the other sides) - say it is directly over the centre of one edge of the base. I have looked at two good solutions on your site, The first, Miter for a pyramid http://mathforum.org/library/drmath/view/54942.html appears to rely on an equal-sided base. Does it? Or is this irrelevant? What is the significance of bisecting the base lines to get m and M? Is it just to get a conveniently known length with a conveniently known angle? If mC and MC were not equal, would mJ and MJ both meet TC at right angles? The 45-degree angle of the line mM cutting across the base - is this significant? Must it be 45 degrees for "dropping a perpendicular" - or is it just an example? This is a beautiful and easy solution, just a bit of ignorance either side of the perfect square base solution makes it hard for me to use. The second solution, Pyramid Construction http://mathforum.org/library/drmath/view/56462.html gives me two problems also. 1. I cannot interpret the lines, the spine and the leaves of the book solution to the pyramid problem. Can you help? (Is it right to consider the lines the base, and the spine the pyramid edge/corner slope?) 2. If you draw your first line from the center of the spine downward at an acute angle (10 degrees from the bottom half of the spine to the line) and the other line opposite it at 30 degrees from the spine, and then close the leaves of the book to 90 degrees, I think you will agree that the angle between the two lines is not 90 degrees, yet the solution suggests this - so I am confused. It looks very elegant and I hope it will solve my weatherproofing construction problem fast if only I can interpret it correctly. Thanks, Simon K.
Date: 09/27/2002 at 11:15:19 From: Doctor Rick Subject: Re: Pyramid internal angles Hi, Simon. Miter for a pyramid http://mathforum.org/library/drmath/view/54942.html does assume that the base is a square and the side triangles are congruent isosceles triangles. Because of the symmetry in the problem, the lines drawn perpendicular to a slanted edge from points equal distances out on the adjacent base edges will meet the slanted edge at the same point; the derivation depends on this. In your case it sounds as if the base is a rectangle so its angles are right angles, but the side triangles are neither isosceles nor congruent. You need a more general solution. Pyramid Construction http://mathforum.org/library/drmath/view/56462.html begins with the same special case (square base with congruent isosceles triangles for the sides). Dr. Mitteldorf, however, discusses the general case. He approaches it by describing a problem that is the reverse of yours: you know the angle between the sides of the pyramid (that's how wide the book is open) and you want to find the base angle (that's the angle between the diagonal lines on the pages). You are quite right in your interpretation: the spine is the slanted edge of the pyramid and the diagonal lines are the edges of the base. Dr. Mitteldorf does not say that the angle between the lines is 90 degrees when the book is open 90 degrees. The angle between the lines is found by his formula cos(x) = cos(a)*cos(b) + sin(a)*sin(b)*cos(c) If you know that angle x is 90 degrees (as in your case), and you know angles a and b (the base angles of the side triangles at the vertex under consideration), then you can solve for cos(c) and then for c itself. The cosine of 90 degrees is 0, so in this case the formula reduces to 0 = cos(a)*cos*(b) + sin(a)*sin(b)*cos(c) which can be solved for cos(c) easily: cos(c) = -cos(a)*cos(b)/(sin(a)*sin(b)) = -1/(tan(a)*tan(b)) c = arccos(-1/tan(a)*tan(b)) If you know or you can find the base angles of the side triangles, then you can find the angle between the triangles using this formula. It will not be 90 degrees unless either tan(a) or tan(b) is infinite, or both are infinite (that is, a and/or b is 90 degrees). As a bonus, here is a way to use geometry to derive the general formula. I'll look at one base vertex of the pyramid. I'll mark off a point on each of the three edges that meet at this vertex (exactly how they are marked off will be described in a moment, but for now, here's the figure so you can see what I'm referring to: The three angles at this vertex are the angle in the base quadrilateral (x = BAC) and the angles in the two adjacent side triangles (a = BAD, and b = CAD). We want to find angle c = BDC. To do this, we can do essentially what Don Mohr (the writer of the second question) did, but for the general case. Mark off point B on one edge of the base, one unit from the vertex A. Drop a perpendicular from point B to the slanted edge, meeting the edge at D. From D draw another line perpendicular to AD and intersecting the second edge of the base at C. Thus angles ADB and ADC are right angles. Now we know enough about the triangles we have constructed to find the lengths of all the sides: AB = 1 BD = sin(a) AD = cos(a) AC = AD/cos(b) = cos(a)/cos(b) CD = AD tan(b) = cos(a)*tan(b) The triangle ABC is not a right triangle in the general case (though I think it is in your problem), so we need to use the Law of Cosines to find BC: BC^2 = AB^2 + AC^2 - 2*AB*AC*cos(x) = 1 + cos^2(a)/cos^2(b) - 2*cos(x)*(cos(a)/cos(b) Now we turn our attention to triangle BCD, which again is not a right triangle, and apply the Law of Cosines: BC^2 = BD^2 + CD^2 - 2*BD*CD*cos(c) 1 + cos^2(a)/cos^2(b) - 2*cos(x)*(cos(a)/cos(b) = sin^2(a) + cos^2(a)*tan^2(b) - 2*sin(a)*cos(a)*tan(b)*cos(c) Multiply through by cos^2(b) to clear the denominators: cos^2(b) + cos^2(a) - 2*cos(x)*cos(a)*cos(b) = sin^2(a)*cos^2(b) + cos^2(a)*sin^2(b) - 2*sin(a)*cos(a)*sin(b)*cos(b)*cos(c) Isolate cos(x) on the left: 2*cos(a)*cos(b)*cos(x) = 2*sin(a)*cos(a)*sin(b)*cos(b)*cos(c) + cos^2(a) + cos^2(b) - sin^2(a)*cos^2(b) - cos^2(a)*sin^2(b) A little trig simplification on the right side gives us 2*cos(a)*cos(b)*cos(x) = 2*sin(a)*cos(a)*sin(b)*cos(b)*cos(c) + 2*cos^2(a)*cos^2(b) Now we can just divide through by (2*cos(a)*cos(b)): cos(x) = sin(a)*sin(b)*cos(c) + cos(a)*cos(b) There's Dr. Mitteldorf's formula, derived by generalizing the method for a symmetrical pyramid. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/
Date: 09/27/2002 at 19:36:17 From: Simon K Subject: Thank you (Angles at an angle - or Mitres on pyramids with non-equilateral sides, and off-centre apexes.) Dear Dr. Rick, Thank you very much for replying so quickly. I really appreciate your site and your effort. It is VERY tough wrestling with problems like this on your own - to most people in ordinary life, this is EXTREMELY obscure stuff. Yet, absolutely essential to resolve a practical problem that is so small - just to join two pieces of wood cleanly! I am replying now (very late at night) because I shall be away for 2 days and feel you deserve a response. I have not yet had time to digest and grasp, let alone apply your solution, so hope to write to you again later when I have. For now - thanks. Simon K.
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