inding the slope to base angle of the virtual pyramid ase 1 igure 1 What we are seeking to find is the measure of angle, or conversely, as triangle is isosceles the two angles at the base will be equal. s 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 is 3 units. Let us proceed to find additional information.
igure 2 We begin by drawing lines connecting point to and. y construction the larger circle has a radius of 2 and since lines and are radii of that circle they also are 2 units in length.
We next draw lines connecting and to, the center of the small, upper circle. igure 3 To solve the triangle it is only necessary to solve one of the base angles since they are both equal. Since our construction is symmetrical about axis we will focus our attention on one side. We can see that we have created isosceles triangle. We know it is isosceles because side = side, since they are both radii of the large circle. 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 is isosceles angle = angle. Remember when designating angles by three letters the vertex letter always goes in the middle. igure 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 and connecting that point back to point. 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 G, but we are going to go ahead and calculate the length of side G for practice. G igure 5 To accomplish this we call upon the Pythagorean Theorem. We know two lengths of triangle G the hypotenuse, which is 2, and the short leg G which is ½, or.5. igure 6 shows the triangle isolated from the rest of the figure with side lengths shown.
2.5 1.936491673 G igure 6 To calculate the length of G 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 G. The process in mathematical symbolism is shown in the equation below. 2 2 2. 5 = 4. 25 = 375. = 1936491673. We now use simple Trigonometry to calculate the angles at and. We will start with angle 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 2 nd key followed by the SI key, activating the inverse of sine, marked in yellow above the SI key as SI 1. Your calculator display should read 14.47751219. This will be the angle at. To find the angle at is easy. We know that triangle G is a right triangle, the angle at G being 90, therefore angles + must equal 90, so to find angle we simply need to know the difference between 90 and angle. With 14.47751219 still in your calculator display press the Subtract key followed by 90. Your display will now read: -75.52248781. ext remove the minus sign by pressing the ± sign immediately to the left of the = sign.
Here are the keystrokes to enter (TI30X).5 2 =.25 in display, press 2 nd SI = 14.47751219 in display, press 90 = 75.52248781 in display. Press ± to remove the negative sign. 2 75.52248781.5 14.47751219 1.936491673 G igure 7 In igure 7 we have angle G 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 is shown below with the two known side lengths of 1. One the next page we will learn how to calculate the angles. 1 1 igure 8
We have calculated angle and found it to be 75.52248781. The angle is going to be exactly the same because Triangle is isosceles. igure 9 The angle is going to be the difference between angle and 180 since angles and are supplementary. Subtracting 75.52248781 from 180 gives us 104.4775122 as the measure of angle. gain, because is isosceles the angles at vertices and are going to be equal and together must sum to the difference between angle and 180 degrees. In other words subtract angle from 180 and divide the remainder by 2 and you will have the measure of the two angles at and.
lso let us recall Proposition 24: The exterior angle of any triangle is equal to the two interior and opposite angles taken together. otice that angle is exterior to triangle. Since angle is 75.52248781 the angles at and taken together must also equal 75.52248781, and, since they are equal in measure they each equal half of 75.52248781, or 37.76124391. elow is triangle completely solved. 104.4775122 1 1 1.58113883 37.76124391 = = 37.76124391 igure 10 otice that the angle at in triangle above is half of the apex angle of the pyramid in igure 9, since triangles and are identical. So we see that 75.52248781 is the measure of apex angle of the pyramid. The two angles at the base of the pyramid triangle are equal to each other and together with the apex angle must sum to 180. oing the math we get 104.4775122 for the sum of the two equal angles at and. ividing 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 52.23875609 52.23875609 igure 9
inding the slope to base angle of the virtual pyramid ase 2 E G H igure 1 To calculate the angles between the sides and base of the Pyramid we must know the length of its sides. t 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 G H igure 2 critical piece of information available to us is the length of H. It is equal to radius, 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 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 and PQ. E P J G H igure 2 Q ext extend the baseline of the pyramid triangle out to meet tangent line. raw a line from point to point H creating triangle JH. We know two side lengths of this triangle: J and H. J is half the Square Root of Three ( 3/2) and H, being equal to is 3.
The Pythagorean Theorem can now be used to calculate the base length of Triangle JH. The calculation is shown below. 2 ( ) 2 3 3 2 9 3 4 8 = = 1 4 = 2.872281323 3/2 3 J 2.872 H igure 4 Knowing that the length of the base of Triangle JH 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. y 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 G and H. This length is 1.372281323. We can use the Pythagorean Theorem to calculate the length of the pyramids sloped side represented by lines G and H 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)
or convenience I am going to call angle G angle α (alpha). We now employ the Tangent unction, which tells us that if we divide side by G, the base of triangle G we will have the value of the Tangent unction unique to the measure of angle α. E P 3 J G α 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.
G = 3 1372281323 = 1262168899.. E 3 J G 1 1.372281323 H K Q With the value of 1.262168899 in your calculator press the Tan 1 key (2 nd T). 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. ompare with ase 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. lso, see the lesson on drawing heptagons. 76.77865488 51.61067256 51.61067256
3 = 1.732050808 51.61067256 or 51 36 38.4 G 1.372281323 H