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-angle 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 quadrilateral ABCD with AB = crd α, BC = crd (180 (α + β)), CD = crd β, AD = 120 (the diameter of the circle). The diagonals are then−AC = crd (180 β) and BD = crd (180 α). Then apply Ptolemy’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. Archimedes’ lemma 2 can be rewritten, using chord notation, in the form (2R)2 crd2(α)+(2R + crd(180 α))2 = . 2 α 2 − crd 2 crd (α) By expanding, using the Pythagorean theorem, 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 Hipparchus’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 line AC at E and the circle at D. Then AD = DC, because these chords subtend equal angles. 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 radius DE will cut AD at G between A and D and will cut DF extended at H. Therefore, sector DEH > triangle DEF , sector DEG < triangle DEA, and triangle DEF : triangle DEA < sector DEH : sector DEG. Since the two triangles have the same altitude, the ratio of their areas 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 vertex 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 latitude 40◦ at noon, the sun is 40◦ below the zenith. 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 angular distance 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 point on the earth 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 symmetry, 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 north 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 longitude 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 area A of an equilateral triangle 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 perpendicular 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− .