Finding the slope to base angle of the virtual pyramid
Case 1
A
F
C
D
M
N
B
Figure 1
What we are seeking to find is the measure of angle AMB, or conversely
ANB, as triangle AMN is isosceles the two angles at the base will be
equal. As always, in trying to solve geometry problems it is helpful to
state what is definitely known. Let’s assume that the radii of the small
circles forming the Vesica are 1, in which case the height of the pyramid
AB is 3 units. Let us proceed to find additional information.
A
F
C
D
M
N
B
Figure 2
We begin by drawing lines connecting point B to C and D. By construction
the larger circle has a radius of 2 and since lines BC and BD are radii of
that circle they also are 2 units in length.
We next draw lines connecting C and D to F, the center of the small,
upper circle.
A
F
C
D
M
N
B
Figure 3
To solve the triangle AMN it is only necessary to solve one of the base
angles since they are both equal. Since our construction is symmetrical
about axis AB we will focus our attention on one side. We can see that
we have created isosceles triangle BCF. We know it is isosceles because
side BC = side BF, since they are both radii of the large circle. CF is equal
to 1, the radius of the small circle, therefore we have an isosceles triangle
with two sides equal to 2 and a base length of 1.
We will add some shading to our triangle of interest to make it more
visible. Since triangle BCF is isosceles angle BCF = angle BFC. Remember
when designating angles by three letters the vertex letter always goes in
the middle.
A
F
C
D
M
N
B
Figure 4
We will now calculate the measure of one of these angles by creating a
right triangle. We do this by bisecting, or finding the midpoint of CF and
connecting that point back to point B. Let’s call our new point G. The
next figure shows the results of our work so far.
We now have enough information to calculate the measure of angle
BCG, but we are going to go ahead and calculate the length of side BG
for practice.
A
C
G
F
D
M
N
B
Figure 5
To accomplish this we call upon the Pythagorean Theorem. We know
two lengths of triangle BGC — the hypotenuse BC, which is 2, and the
short leg CG which is ½, or .5. Figure 6 shows the triangle isolated from
the rest of the figure with side lengths shown.
C
2
B
1.936491673
.5
G
Figure 6
To calculate the length of BG proceed as follows: Square the hypotenuse
and from that result subtract the square of the short leg. Take the square
root of the remainder and that will be the length of BG. The process in
mathematical symbolism is shown in the equation below.
22 −.52 =
.
4−.25 = 3.75 = 1936491673
We now use simple Trigonometry to calculate the angles at B and C.
We will start with angle B by using the Sine function (opposite divided by
hypotenuse).
We first divide .5, the length of the opposite side, by 2, the length of
the hypotenuse, which gives us .25. This is the Sine function for a unique
angle found on your calculator by pressing the 2nd key followed by the SIN
key, activating the inverse of sine, marked in yellow above the SIN key as
SIN—1. Your calculator display should read 14.47751219°. This will be the
angle at B. To find the angle at C is easy. We know that triangle BGC is a
right triangle, the angle at G being 90°, therefore angles B + C must equal
90°, so to find angle C we simply need to know the difference between 90°
and angle B. With 14.47751219 still in your calculator display press the
Subtract key followed by 90. Your display will now read: -75.52248781.
Next remove the minus sign by pressing the ± sign immediately to the left
of the = sign.
Here are the keystrokes to enter (TI30XA) .5 ÷ 2 = .25 in display, press
2nd SIN = 14.47751219 in display, press – 90 = —75.52248781 in
display. Press ± to remove the negative sign.
C
2
75.52248781°
14.47751219°
B
.5
G
1.936491673
Figure 7
In Figure 7 we have angle BGC completely solved. We say a triangle is
completely solved when we know the measure of all three angles and the
lengths of all three sides.
Isosceles Triangle AFC is shown below with the two known side lengths
of 1.
One the next page we will learn how to calculate the angles.
F
1
A
1
C
Figure 8
We have calculated angle BCF and found it to be 75.52248781°. The
angle BFC is going to be exactly the same because Triangle BCF is
isosceles.
A
F
C
M
D
B
N
Figure 9
The angle AFC is going to be the difference between angle BFC and 180°
since angles AFC and BFC are supplementary. Subtracting 75.52248781°
from 180° gives us 104.4775122° as the measure of angle AFC. Again,
because AFC is isosceles the angles at vertices A and C are going to be
equal and together must sum to the difference between angle F and 180
degrees. In other words subtract angle F from 180 and divide the
remainder by 2 and you will have the measure of the two angles at A and
C.
Also let us recall Proposition 24: The exterior angle of any triangle is
equal to the two interior and opposite angles taken together. Notice
that angle BFC is exterior to triangle AFC. Since angle BFC is
75.52248781° the angles at A and C taken together must also equal
75.52248781°, and, since they are equal in measure they each equal half
of 75.52248781°, or 37.76124391°.
Below is triangle AFC completely solved.
104.4775122°
F
1
1
37.76124391° = A
1.58113883
C = 37.76124391°
Figure 10
Notice that the angle at A in triangle AFC above is half of the apex
angle of the pyramid in Figure 9, since triangles AFC and AFD are
identical. So we see that 75.52248781° is the measure of apex angle A of
the pyramid. The two angles at the base of the pyramid triangle AMN are
equal to each other and together with the apex angle must sum to 180°.
Doing the math we get 104.4775122 ° for the sum of the two equal
angles at M and N. Dividing this by 2 gives us 52.23875609° for the slope
to base angle of this virtual pyramid cross section.
We have demonstrated that in this case the angle is slightly larger than
the measured slope to base angle of the Great Pyramid of Khufu of
51.85°
Here is the pyramid with angles shown.
75.52248782°
A
C
M
D
52.23875609°
52.23875609°
B
N
Figure 9
Finding the slope to base angle of the virtual pyramid
Case 2
E
C
M
A
G
D
B
H
Figure 1
F
To calculate the angles between the sides and base of the Pyramid we must
know the length of its sides. At this point we do not have that information.
In solving problems of this kind we must first determine what we do know.
We assume that the radii of our two small generating circles is 1 unit.
Procedure
E
C
M
A
G
D
B
H
Figure 2
F
A critical piece of information available to us is the length of AH. It is equal
to radius AB, which is 3, the radius of the large circles whose intersecting
arcs form the enclosing Vesica.
If the radius of the small circles is 1 then the width of the small Vesica is also
1. That means that the length of the small Vesica measured by CD is the
Square Root of 3. To proceed with the calculation place vertical lines
tangent to the two arcs of the large Vesica as shown below. We call the
tangent lines MN and PQ.
M
E
P
C
A
J
B
G
D
H
Figure 2
N
F
Q
Next extend the baseline of the pyramid triangle out to meet tangent line
MN. Draw a line from point A to point H creating triangle AJH. We know
two side lengths of this triangle: AJ and AH. AJ is half the Square Root of
Three (√3/2) and AH, being equal to AB is 3.
The Pythagorean Theorem can now be used to calculate the base
length of Triangle AJH. The calculation is shown below.
(3 )
2
2
−  3 2  = 9 − 3 4 = 8 1 4 = 2.872281323


A
√3/2
J
3
2.872
H
Figure 4
Knowing that the length of the base of Triangle AJH is 2.872281323
we can subtract that number from 3 to get the length of the small
increment JG. Refer to the next figure to see that this difference is
repeated twice in both JG and HK. By subtracting the number
2.872281323 from 3 we obtain the length of JG which is .127718677.
If we double this number to account for the same increment on the
other side, then subtract it from 3 we have the total length of the
pyramids base, which is 2.744562647. If we now divide this number
by 2 we have the base length of both right triangles CDG and CDH.
This length is 1.372281323. We can use the Pythagorean Theorem to
calculate the length of the pyramids sloped side represented by lines
CG and CH if we should so desire. However, to get the angle measure
of the slope to base, which is what we are after, we do not need to
know that length. We merely employ the Tangent function of
opposite divided by adjacent. (Rise over run)
For convenience I am going to call angle CGD angle α (alpha). We now
employ the Tangent Function, which tells us that if we divide side CD
by GD, the base of triangle CGD we will have the value of the Tangent
Function unique to the measure of angle α.
E
M
P
C
√3
A
J
G
N
α
D
F
B
H
K
Q
Then by using the Tan—1 key we can convert that function value into its
corresponding angle. The next page shows the equation for calculating
the value of Tan α.
CD
3
=
= 1262168899
.
GD 1372281323
.
E
C
A
B
√3
J
G
N
1
1.372281323
D
F
H
K
Q
With the value of 1.262168899 in your calculator press the Tan—1 key (2nd
TAN). The resulting value is the measure of angle α, which =
51.61067256° slightly less than the measured slope to base angle of the
Great Pyramid of Khufu, which angle is 51.85°.
Here is the virtual pyramid with all angles shown. Compare with Case One.
Which pyramid is closer to the measured angle of Khufu’s Pyramid of
51.85°? There are other ways of generating the profile of the Great Pyramid
which are more accurate than these two cases. One method involves the
use of the Golden Section and will be taught in an upcoming class. Also, see
the lesson on drawing heptagons.
76.77865488°
51.61067256°
D
51.61067256°
C
√3 = 1.732050808
51.61067256° or
51° 36’ 38.4”
G
D
1.372281323
H