CHAPTER 5

1. We first need to calculate crd 120 . We get crd 120 = √4R2 R2 = √3R = √3 60 = ◦ ◦ − · 103; 55, 23. Then the half- formula gives crd 30 = R(2R crd 120 ) = 31; 03, 30. ◦ − ◦ Next, crd150 = 4R2 crd2 30 = 115; 54, 40, so crdq 15 = R(2R crd 150 ) = ◦ − ◦ ◦ − ◦ q 1 ◦ q 15; 39, 47. Similarly, crd 165◦ = 118; 58, 25 and crd 7 2 = 7; 50, 54. 1 ◦ 2. We have calculated the chords of 120◦, 150◦, and 165◦ in exercise 1. Thus, crd 172 2 = 4R2 crd2(7 1 ◦) = 119; 42, 28. − 2 3.q Since 672 = 4489, √4500 = 67; x,y. Now 4500 672 = 11; divide 11 60 by 2 67 to get x = 4. So √4500 = 67; 4,y. Now 4500 (67;− 4)2 = 2; 3, 44 = 7200 +· 180 + 44· = 7424 seconds. Then 7424 2(67; 4) = 55 to the− nearest integer. So √4500 = 65; 4, 55. ÷ 4. Use a ABCD with AB = crd α, BC = crd (180 (α + β)), CD = crd β, AD = 120 (the of the ). The diagonals are then−AC = crd (180 β) and BD = crd (180 α). Then apply ’s theorem. − − 5. 120 crd(72 60) = crd(72) crd(120) crd(60) crd(108). So 120 crd(12) = 70; 32, 3 103; 55, 23 − 60 97; 4, 56 = 1505; 11−, 34. It follows that crd(12) = 12; 32, 36. Then× − × crd(168) = 4 602 crd2(12) = 119; 20, 33. Thus crd(6) = 60(2 60 crd(168)) = × − × − q 1 q 3 6; 16, 49. Similarly, crd(3) = 3; 8, 29; crd(1 2 ) = 1; 34, 15; and crd( 4 ) = 0; 47, 7. 6. ’ lemma 2 can be rewritten, using notation, in the form

(2R)2 crd2(α)+(2R + crd(180 α))2 = . 2 α 2 − crd 2 crd (α)   By expanding, using the , and simplifying, this result can be put into the form α R crd2(α) crd2 = . 2 2R + crd(180 α)   − Multiplying numerator and denominator of the right side by 2R crd(180 α) and simplifying then gives ’s theorem. − − 7. Let crd α = AB and crd β = BC. First, bisect the angle at B, and extend the angle bisector until it meets the AC at E and the circle at D. Then AD = DC, because these chords subtend equal . Also, since BE bisects the angle at B, we have by Elements VI–3 that AE : EC = AB : BC. Since AB < BC, we also have AE < EC. Next, drop a perpendicular DF from D to AC. Because D bisects arc ADC, it follows that F is the midpoint of AC. Then a circle of center D and DE will cut AD at G between A and D and will cut DF extended at H. Therefore, sector DEH > DEF , sector DEG < triangle DEA, and triangle DEF : triangle DEA < sector DEH : sector DEG. Since the two have the same , the ratio of their is the same as the ratio of their bases. Also, the ratio of the sectors is the same as the ratio of the corresponding angles with at D. Thus EF : EA < 6 EDH : 6 EDG. If we add 1 to each side of the inequality, we get the new inequality AF : EA < 6 GDH : 6 EDG. 27 28 Chapter 5

If we double the numerators, we then get AC : EA < 6 ADC : 6 EDG, or, subtracting 1 from both sides, that EC : EA < 6 EDC : 6 EDG. The left side of this inequality is equal to BC : AB, while the right side is equal to the corresponding ratio of arcs, namely β : α. Thus we have the final result that crd β : crd α<β : α, as desired.

8. At the vernal equinox at 40◦ at noon, the sun is 40◦ below the . Using figure 5.19, CF = crd 80◦ = 77; 8, 5 and CE = crd 100◦ = 91; 55, 31, so the shadow = 60 77; 8, 5 = 50; 21. 91;55,31 · 9. Note that  is the latitude where the sun is directly overhead at noon on the summer solstice. The between the noon altitudes of the sun at the summer and winter solstice is, given the assumption that at any given time the sun’s rays to every on the are parallel to each other, equal to the angle between the sun at noon on the summer solstice and the sun at noon on the winter solstice, as viewed from the center of the earth. And this angle, by Figure 5.40, is twice . 1 ◦ 10. At noon on the summer solstice at latitude 36◦, the sun is 12 2 below the zenith, so the 1 ◦ 1 ◦ length of the shadow is 60 tan 122 = 13; 18, 6. At the winter solstice, the sun is 59 2 1 ◦ below the zenith and the shadow length is 60 tan 592 = 101; 51, 35. 11. When λ = 90◦, then δ = 23◦510 and α = 90◦. When λ = 45◦, we have sin δ = sin(23◦510) sin(45◦) and δ = 16◦370. Also tan α = cos(23◦510) tan(45◦), so α = 42◦270. By , the values for the declination at 270◦ and 315◦ are the negatives of the values at 90◦ and 45◦, respectively.

12. To calculate ρ(60◦, 45◦), we note that if λ = 60◦, then δ = 20◦300 and α = 57◦440. Since sin σ = tan δ tan 45◦, we have σ = 21◦570 and ρ = α σ = 35◦470. If λ = 90◦, then δ = 23 51 and α = 90 . So σ = 26 14 and ρ = α σ =− 63 46 . ◦ 0 ◦ ◦ 0 − ◦ 0 13. L(λ, φ) = 180◦ + 2σ(λ, φ). When λ = 60◦ and φ = 36◦, we calculate that sin σ = tan δ tan φ = tan(20◦300) tan(36◦); so σ = 15◦460 and L = 211; 32, which corresponds to 14 hours, 6 minutes. Therefore, sunrise is 7 hours, 3 minutes before noon, or 4:57 a.m. and sunrise is at 7:03 p.m.

14. If the length of day is 15 hours when λ = 90◦, then 180◦ + 2σ(90◦,φ) = 225◦. Therefore ◦ 0 sin(22 30 ) σ = 22◦300 and, since sin σ = tan δ tan φ, we have tan φ = tan(23◦510) , so φ = 40◦530. ◦ 0 sin δ sin(23 51 ) Also, since sin β = sin(90 φ) = sin(49◦70) , we calculate that β = 32◦200. That is, the − sunrise occurs at 32◦200 of east and sunset at the same angle north of west. By symmetry, the positions of sunrise and sunset at the winter solstice occur 34◦200 south of east and south of west respectively. 1 ◦ 15. The expression tan δ tan φ will be greater than 1 for δ = 23 2 when .4348 tan φ > 1, or Mathematical Methods in Hellenistic Times 29

1 ◦ when tan φ> 2.2998, or when φ> 66 2 . When that occurs, the formula for L no longer makes sense. Since when tan δ tan φ = 1, we know that L = 360◦ or 24 hours, it follows that the sun does not set at all on the summer solstice when the latitude is greater than 1 ◦ 66 2 . 16. If λ = 45 , the δ = 16 37 ; so SZ = φ δ = 45 16 37 = 28 23 . Similarly, if λ = 90 , ◦ ◦ 0 − ◦ − ◦ 0 ◦ 0 ◦ then δ = 23◦510 and SZ = 21◦90. sin δ 17. The sun is directly overhead at noon at latitude 20◦ when δ = 20◦. Since sin λ = sin 23.5◦ , we find that λ = 59◦. This value for the of the sun occurs at approximately 60 days after the spring equinox and 60 days before the fall equinox, or at approximately May 20 and July 21.

18. The maximal northerly sunrise point occurs when λ = 90◦ and therefore when δ = 23◦510. When φ = 36◦, we calculate that sin δ sin 23 51 sin 23 51 sin β = = ◦ 0 = ◦ 0 = 0.4998, sin(90 φ) sin(90 36) sin 54 − − ◦ ◦ so β = 29◦590. sin δ 19. When φ = 75◦, we need to find δ so that sin 15◦ = 1. Clearly, δ = 15◦, and since sin 15◦ = sin 23◦510 sin λ, it follows that λ = 39◦480. This value occurs approximately 40 days after the vernal equinox, or about April 30. √3 2 20. The exact A of an of side s is A = 4 s . Thus the Roman surveyor has approximated √3/4 by 1/3 + 1/10. This is the same as approximating √3 by 4/3 + 4/10 = 26/15 1.733. ≈ 21. Suppose points A and B are on the opposite bank of the river. One possible method to determine the distance from A to B is the following: First determine the distance across the river from a point C on your side (using the method in the text). For simplicity, we will assume that line AB is to line AC. Choose a point D on the extension of line AC and construct a line CF perpendicular to AC. Next, determine the point E on CF where the line of sight from D to B intersects CF . Since triangles ABD and CED are similar, and since the distances AD and CE are known, the distance AB can be calculated. 22. In Fig. 5.34, we will take a = 10, b = 7, and c = 4. The method based on Elements 2 2 2 a +b c II–12, 13 gives us that CD = 2a− . Thus CD = 6.65, h = AD = 2.19, and the area 1 21 = 2 ah = 10.93. To use Heron’s formula, we first calculate s = 2 = 10.5 and then find that the area is √10.5 6.5 3.5 0.5= √119.4375 = 10.93. · · · 23. Divide the pentagon into five equal isosceles triangles with a, side r, and altitude 2 p. As on page 88 of the text, a = r 10 2√5. Since p2 = r2 a , a computation 2 − − 2 shows that q   3+ √5 p = v a u20 4√5 u − 5 t and therefore, since A5 = 2 pa, that 5 5 + 2√5 A = a2. 5 2s 20 30 Chapter 5

5 2 This value is roughly approximated by Heron’s value A5 = 3 a . 8 24. Given that the radius r of the circle is 7 a, where a is the side of the inscribed heptagon, it follows that the area of the heptagon is A = 7 1 207 a2 = 3 √23a2. Heron has therefore · 2 196 4 3 √ 43 √ 43 q approximated 4 23 by 12 or 23 by 9 . √ 1 8 17 25. Start with the approximation 8 = 3. Then 2 (3 + 3 )= 6 . 26. Since the regular octahedron is made up of two pyramids, each with base of side a, the volume is double the volume of one of these pyramids. But the volume of a pyramid 1 2 is 3 hB, where h is the height and B the area of the base. In this case, B = a and h is √3 a one leg of a whose is 2 a and whose other leg is 2 . Therefore, √2 1 √2 2 √2 3 h = 2 a, and the volume of the octahedron is given by V = 2 3 2 aa = 3 a , as asserted by Heron. · 27. The actual distance between Alexandria and Syene (modern Aswan) is approximately 850 km or 520 miles. Aswan is located at 24◦50 N latitude and 32◦560 E longitude, while the coordinates of Alexandria are 31◦120 N and 29◦540 E. Thus Aswan is not on the Tropic of Cancer, but is approximately 150 north of it (using Ptolemy’s value). Also, Alexandria and Aswan do not have the same longitude; Aswan is approximately 3◦ east of Alexandria. On the other hand, the difference in degrees of the of the two 1 ◦ cities is 7◦70, which is very close to ’ value of 7 5 . If there were 5000 stades between the two cities, then a stade would be approximately 0.17 km = 170 meters. 28. Let the total length of a parallel at latitude α be Cα and the of the earth at the be C. Let the radius of the latitude circle at latitude α be rα and the radius of the earth be r. Then Cα : C = rα : r = cos α, a relationship easily seen by constructing the right triangle in a cross section of the earth one of whose sides is rα and whose hypotenuse is r. Thus, Cα = C cos α. 29. Since the ratio of a at latitude α to a degree at the equator is as cos α (by 5 ◦ exercise 28), we just need to check approximations to cos α. cos 23 6 = 0.9147, while 7 5 ◦ 5 4 12 :5=0.9167. cos 16 12 = 0.9592, while 4 6 :5=0.9667.