Difference between revisions of "The Seqed and Modern Trigonometry Exploration"

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[[category:Egyptian Mathematics Exploration]]

Revision as of 08:29, 24 January 2009

The Seqed and the Cotangent

Several of the problems in the Rhind papyrus compute the so called seqed of a pyramid. The Egyptians computed the ratio (half the base)/height.

AreaTriangle.gif

In the image of the triangle here the seqed is actually the run over the rise of the smaller right triangle that makes up the left half of the image. Note that in this case the seqed measures the inverse of the slope (which would be the rise over run). Recall that for a right triangle we can define several trigonometric functions. the best known of those are:

  • sin(x) = opposite / hypothenuse
  • cos(x) = adjacent / hypothenuse
  • tan(x) = opposite / adjacent
  • cotangent(x) = adjacent / opposite


So for the small right triangle that takes up half of the larger triangle - which can be thought of as a cross section of a pyramid - we can look at the cotangent and notice that it is exactly the seqed computed by the ancient Egyptians.

cotangent(x) = adjacent / opposite = half the base / height = seqed

So what exactly is the seqed (or if you prefer the cotangent) measuring? It gives a ratio of the base versus the height and hence gives a measuer of the steepness of the triangle. And because the triangle really represents the pyramid, it measures how steep the pyramid is.

Note that a very steep pyramid would have a relatively great height in comparison to the size of the base. So smaller seqeds correspond to higher pyramids, which in turn correspond to steeper pyramids

Example:
Rhind Papyrus 56: Example of reckoning a pyramid 360 cubits wide and 250 cubits high. Cause thou that I know the seked of it. You are to take half of 360; It becomes 180. You are to reckon with 250 to find 180. Result: 1/2 + 1/5 + 1/50. (slight adjustment of the text to make it easier to read)

What the Egyptian scribe is doing here is showing how to compute the seqed.
The first part is straight forward:
You are to take half of 360; It becomes 180. We need to use half the base length, so we divide the given measurement by 2.
You are to reckon with 250 to find 180. Result: 1/2 + 1/5 + 1/50. This part means divide 250 into 180 and the answer is 1/2 + 1/5 + 1/50, which we in modern days would simplify to 18/25, or if we use a calculator would be represented as .72.

Question 1:
Compute problem 58 from the Rhind papyrus: In a pyramid whose height is 93 1/3 [cubits] make known the seqed of it when its base side is 140 [cubits].

Question 2:
Solve problem 57 from the Rhind papyrus: If the seqed is 3/2 and the base is 280 [cubits], what is the height of the pyramid?

Question 3:
Solve problem 59(a) from the Rhind papyrus: In a pyramid whose base side is 12 [cubits] and whose altitude is 8 [cubits]; what is its seqed?

Real Pyramids, right triangles and their Seqeds

One theory for how the Egyptians planned the construction of their pyramids and maintain a constant slope is that they used right triangles. The idea is that they may have constructed wooden right triangles that could have been used as a guide during the construction of the monument. Egyptians may have known about perfect triangles from experimenting with their version of a measuring tape: knotted cords. It may have been noted for instance that if one has pieces of ropes of length 3, 4, and 5 that they form a right triangle. There is no real evidence that the Egyptians knew this back in the day of the pyramid builders, but it seems reasonable that they may have noticed that there are some perfect triangles out there.

Question 4:
Below are some Pythagorean triples. Compute the seqed - in this case b/a - for each of the triangles.

Triple (a, b, c) Corner Angle Seqed (b/a)
( 3, 4, 5) <math>53.1^o</math>
( 5, 12, 13) 67.4<math>^o</math>
( 7, 24, 25) <math>73.7^o</math>
( 8, 15, 17) <math>61.9^o</math>
( 9, 40, 41) <math>77.3^o</math>
(11, 60, 61) <math>79.6^o</math>
(12, 35, 37) <math>71.1^o</math>
(13, 84, 85) <math>81.2^o</math>
(16, 63, 65) <math>75.6^o</math>
(20, 21, 29) <math>46.4^o</math>
(28, 45, 53) 58.1<math>^o</math>
(33, 56, 65) <math>59.5^o</math>
(36, 77, 85) <math>64.9^o</math>
(39, 80, 89) <math>64.0^o</math>
(48, 55, 73) 48.9<math>^o</math>
(65, 72, 97) <math>47.9^o</math>


Question 5:
For the following pyramids do the following:
(i) Compute the Seqed if possible.
(ii)Could the pyramid have been constructed using a technique involving right triangles?

Pyramid Height base Seqed of the pyramid Slope Possible right triangle
Great Pyramid (Pharaoh Khufu) 146.6 230.3 51.8 degrees
Pharaoh Khafre 143.5 215.25 .75 53.1 degrees 3-4-5 Triangle
Pharaoh Menkaure 66.45 103.4 .79 51.3 degrees Maybe 3-4-5 Triangle?
Pharaoh Userkaf 49 73.3 53 degrees
Pharaoh Sahure 48 78.5 50.5 degrees
Queen Khentkawes 72 25 52 degrees
Pharaoh Djedkare 52.5 78.75 52 degrees
Djedkare's Queen 21 41 ca 45 degrees
Pharaoh Teti 52.5 78.5 53.1 degrees
Teti's Queen (Iput) 21 21 63 degrees