Contributors: He a v e n l y Ma t h e ma t i c s T h ef o r g o t t e nar to f S p h e r i c a l T r i g o n o me t r y Gl e nva nbr umme l e n Solutions Manual Edition: 25 April 2013 Editors: Adam W. Parke, Thomas B. Turner, Glen Van Brummelen Sean Holt Adam Parke Tom Turner
Contents Preface Errata for Heavenly Mathematics List of Contributors i ii iii 1 Chapter 1 Solutions 1 2 Chapter 2 Solutions 2 3 Chapter 3 Solutions 5 4 Chapter 4 Solutions 8 5 Chapter 5 Solutions 11 6 Chapter 6 Solutions 15
Preface Welcome to the solutions manual for the book Heavenly Mathematics. Its purpose is to aid the reader through some of the exercises from the book. This manual assumes a solid basis in planar trigonometry as well as fluency in algebra; readers are still strongly encouraged to attempt the problems themselves before resorting to this manual. This is a student-led project and an online book, which we will update frequently as we add solutions to more problems. Students and former students of the course Spherical Trigonometry at Quest University Canada are writing the solutions. Because of this, the solutions might be updated in any order. Solutions from readers are encouraged and may be submitted to the editors for possible inclusion in this manual. Not all of the exercises in the book will have solutions in this manual. Heavenly Mathematics is being used as a textbook for a course; thus some problems must remain unsolved here so they can be assigned as homework. The exercises that will never have solutions here are: Chapter 1 : 1, 2, 6, 7, 13, 14 Chapter 2 : 3, 4, 5, 6, 7, 8, 10 Chapter 3 : 3, 4, 5, 6, 7 Chapter 4 : 1, 2, 7, 8 Chapter 5 : 3, 4abc, 5a, 6a, 7, 8, 12, 13 Chapter 6 : 1ab, 6, 8, 9, 10, 11, 12 We hope you find these solutions helpful. If you have any suggestions or corrections please let us know. -Adam Parke and Thomas Turner, editors (Thomas.Turner@questu.ca)(Adam.Parke@questu.ca) i
Errata for Heavenly Mathematics Page 40: Exercise #12 refers to a spherical polygon with n sides (each 180 ). It should read: (each < 180 ). Page 71: Exercise #7 should refer to exercise #6 from chapter 3, and exercise #8 should refer to exercise #7 from chapter 3. Page 80: The equation should be c 2 = a 2 + b 2 + a2 b 2 2 c 2 = a 2 + b 2 a2 b 2 2. Page 105: The phrase...and so c = 28, equal to 1680 nautical miles or 1933 statute miles. should read...and so c = 56, equal to 3360 nautical miles or 2866 statute miles. ii
List of Contributors Contributors are identified in the solutions by their initials. Sean Holt [SH] Adam Parke [AP] Thomas Turner [TT] iii
1 Chapter 1 Solutions 12) Label the intersection point on AB as D. Draw a horizontal line rightward from C and drop a perpendicular line from B onto it defining Y. Since ACY is a right angle, ACB = 85 ; and ACD = 90 43.867 = 46.133. Then DCB = ACB ACD = 38.867. Furthermore since we know the altitude of the mountain and the altitude of the airplane, we have AC = 8460 ft 4135 ft = 4325 ft, thus sin 43.867 = CD CD = 2997.167 ft. 4325 Also cos 38.867 = 2996.167 CB = 3848.36 ft. CB Lastly, sin 5 = BY BY = 335.4 ft. 3848.36 Then the height of the mountain B is 4135 ft + 335.4 ft = 4470.4 ft. [SH] 1
2 Chapter 2 Solutions 1)a) To find eccentricity EC, we first find the lengths of the arcs in terms of degrees. For spring, and summer is 94.5 s 92.5 q = 365.25 360 s = 93.142, = 365.25 360 q = 91.170. Figure 2.1: From figure 2.5 in Heavenly Mathematics with the apogee marked η. In Figure 2.1, arc length for spring, s, is 93.17 = 90 + θ + φ. The arc length for summer, q, is 91.142 = 90 φ + θ. Simplifying, we find: 3.142 = θ + φ (1) Substituting (2) into (1) gives θ = 1.17 + φ. (2) 3.142 = 1.17 + φ + φ, simplifying, we find φ = 0.986. Using this value in (1) gives θ = 2.156. Now we find the lengths of the sides of the rectangle containing C and E. Because CA = 1, sin θ = CD = 0.03762, and 2
3 The Pythagorean Theorem tells us sin φ = ED = 0.01721. CE = ED 2 + CD 2 = 0.000296 + 0.00142 = 0.04137. b) Extend EC to the edge of the circle at η furthest from Earth. The angle between Cη and Cx is the same as CED, so The arc > ηa is sin CED = CD CE =.9131 CED = 65.42. CED + θ = 65.42 + 2.156 = 67.572. [AP ], [T T ] 2) The lune s surface area is θ (Surface area of unit sphere). 360 Since the surface area of the unit sphere is 4πr, θ 360 4πr2 = θ4πr2 360 = θπr2. [T T ], [AP ] 90
4 2. CHAPTER 2 SOLUTIONS 12) An n-sided spherical polygon can be divided into n triangles by foining vertices to a central point within the polygon. Since we know that the total angle measurement in a spherical triangle must be > 180, for n triangles the angle sum will be > 180n. We are only after the angles of the polygon, not the total angles of the triangles, so we remove the angles surrounding the central point (which sum to 360 ) from the total sum. This leaves the angle sum > 180n 360. [AP],[TT] Figure 2.2: An n-sided polygon on a sphere.
3 Chapter 3 Solutions 1) a) Assign λ = 42 and use Menelaus s Theorems. Figure 3.1: Similar to Figure 3.3 from Heavenly Mathematics, used for calculations. To find the right ascension, apply Theorem A to Figure 3.1: This gives Simplifying, we have sin AB sin A sin(90 α) sin α = = sin BN sin CN sin 90 sin(90 ɛ) sin C sin. sin(90 λ). sin λ cot α = 1 cot λ tan α = cos ɛ tan λ. cos ɛ Using known values we find α = tan 1 (cos 23.4 tan 42 ) = 39.57. 5
6 3. CHAPTER 3 SOLUTIONS To find the declination, we apply Theorem B to Figure 3.1: This gives sin B sin BA sin 90 sin(90 α) = Simplifying, we have 1 cos α = 1 cos λ Using known values, we find b) Using Menelaus Theorem B Solve for λ δ = cos 1 ( cos 42 cos 39.57 sin 90 sin(90 α) = sin BC sin N = sin C sin NA. sin 90 sin(90 λ) sin(90 δ). sin 90 cos δ cos δ = cos λ cos α. ) = 15.41. [T T ], [AP ] sin 90 sin(90 λ) sin(90 δ) sin 90 λ = cos 1 (cos α cos δ). Substituting α = 126.31 and δ = 19.22 we find λ = 123.99, and from Appendix A we know that this value corresponds to July 28 th. 2) Using the steps from page 54 of Heavenly Mathematics we solve for δ, η, n, θ. We are given ɛ = 23.4, φ = 49.3 and from Appendix A we find λ = 58.3. a) To solve for δ we use conjunction on the Menelaus configuration NGZM to determine sin δ = sin λ sin ɛ δ = sin 1 (sin 58.3 sin 23.4 ) = 19.75. b) To solve for η we use conjunction on the Menelaus configuration NHQME sin η = sin δ cos φ. Using this we subsitute in our values and solve for η ( ) sin 19.75 η = sin 1 = 31.21. cos 49.3
7 Figure 3.2: Figure 3.9 from Heavenly Mathematics. c) To solve for n we use Menenlaus configuration NHQME with conjunction, but the arc assignment changed sin MQ = cos η cos δ where MQ = 90 n. We find ( ) cos 31.21 n = cos 1 = 24.67. (1) cos 19.75 d) We find MZ using Menelaus configuration NGZM and altering the arc assignemnt from a) Substituting; we find Combining (1) and (2) we find sin MZ = cos λ cos δ. ( ) cos 58.3 MZ = sin 1 = 33.93. (2) cos 19.75 θ = ZQ = MQ MZ = (90 24.67 ) 33.93 = 31.4. [AP ], [T T ]
4 Chapter 4 Solutions 3) F D E A B C Figure 4.1: Abū Nasr s figure (Figure 4.1 in Heavently Mathematics). Apply Menelaus conjunction to Figure 4.1; we get sin AC sin CB = sin EA sin DE sin F D sin F B. Apply Menelaus disjunction to the same figure; we get Solving both results for sin CB, sin DE sin CB sin DE = sin F C sin EF sin CB sin AB = sin F C sin EF sin AB sin AD sin DE sin AD. Setting solved conjunction and disjunction equal, we get = sin AC sin EA sin F B sin DF. sin F C sin EF sin AB sin AD sin AC = sin EA sin F B sin DF (1) 8
9 Since F BC and ACE from Figure 4.1 both have two 90 angles, Substituting (2) into (1) and simplifying gives us AC = F C = EA = F B = 90. (2) sin DF sin EF sin AD =. [T T ], [AP ] sin AB 6) Figure 4.2: Indian approach to finding arcs of the ecliptic (Figure 4.4 in Heavenly Mathematics). In Figure 4.2, AO = α, OD = δ, and COK = ɛ. Therefore, and We also know ED COK, so We also found on page 65 that sin α = ED OD, (1) sin δ = D, cos δ = OD, (2) sin ɛ = CK, cos ɛ = OK. (3) E OC = D CK = ED OK. (4) sin δ = sin λ sin ɛ. (5) Substituting (2) into (5) gives D = sin λ sin ɛ. Using this along with (3) and (4), we find D CK = sin ɛ sin λ sin ɛ = sin λ = ED OK ED = sin λ cos ɛ.
10 4. CHAPTER 4 SOLUTIONS Combining this with (1) and (2), we find the Indian theorem for the right ascension: sin α = sin λ cos ɛ. [AP ], [T T ] cos δ 9) a) On construction P GCBF, we know > PB = > PC = 90. We are also given > BC = 27.37, > GC = 33.58. From these values we can use the Rule of Four Quantities to find > FG: sin > FG sin > BC = sin > 90 GC sin > sin > FG = cos > GC sin > BC. PC Substituting in known values: sin > FG = cos 33.58 sin 27.37 = 0.880117; > FG = 22.52. b) Using construction EF P NH we now solve for > PE. In Figure 4, > EF = > HE = 90. We also know that > PN = 90 > PG. Because > PG = 90 > GC, > PN = > GC. We also know > PE = 90 + > FP, so sin(90 > FG) = sin > PN sin 90 sin > PE cos> sin > PN FG = sin (90 + > FP) cos> FP = sin > PN cos > FG. Substituting known values gives cos > sin 33.58 FP = cos 22.52 = 0.59876; > FP = 53.21 > PE = 90 + 53.21 = 143.21. From figure 6, > PB = > EF = 90 ; to determine > ME we need to find > FB and > FM. First find > FB > FB = > PB > PF = 36.79 We can use > FB and > MB = 21.67 to determine > FM: > FM = > FB > MB = 15.12. Since > ME = 90 - > FM we have > ME = > EF > FM = 74.88. c) To find > MD, apply the Rule of Four Quantities on construction EMF DH: sin > MD sin > FH = sin > ME sin > EF > MD = 63.1. d) To find the qibla simply apply the spherical Law of Sines to MP G. From figure 6, GP M = > BC = 27.37, > MG = 90 > MD, and > PM = > PF + > FM = 68.33 ; so sin > MG sin GP M = sin > ( ) PM sin(68.37 ) sin(27.37 ) MGP = sin 1 = 70.83. [T T ], [AP ] sin MGP cos(63.1 )
5 Chapter 5 Solutions 1) Figure 5.1: Tetrahedron ODEF formed between spherical triangle ABC and center point O (similar to Figure 5.2 in Heavenly Mathematics). From the above figure, sin b = EF OE = EF ED ED OE = cot A tan a. cos A = EF DF = EF OF OF = tan b cot c. DF cos c = OF OD = OF ED ED = cos b cos a. [AP ], [T T ] OD 2) To derive cos c = cot A cot B we start with cos c = cos a cos b. We solve sin a = tan b cot B and sin b = tan a cot A for cos a, and cos b, sin a = tan b cot B cos b = sin b sin a cot B sin b = tan a cot A cos a = sin a cot A. sin b 11
12 5. CHAPTER 5 SOLUTIONS Substituting our new values for cos a and cos b, we arrive at cos c = cot A cot B. To derive cos A = cos a sin B we start with cos A = tan b cot c, which we rewrite as cos A = sin b sin c cos c cos b. (1) Using the identities sin b = sin B sin c and we solve for sin b sin c cos c and cos b. This gives sin b sin c Substituting (2) into (1) we have cos c = cos a cos b = sin B and cos c cos b cos A = cos a sin B. = cos a. (2) To derive cos B = cos b sin A we rewrite the formula cos B = tan a cot c as follows: cos B = sin a sin c cos c cos a. (3) Using the identities and we solve for sin a sin c and cos c cos a : cos c = cos a cos b sin a = sin A sin c sin a sin c Substituting (4) into (3), we have = sin A and cos c cos a cos B = cos b sin A. [T T ], [AP ] = cos b. (4)
4) d) Given the data a = 69.72 and c = 78.42, we can find the rest of the right triangle, starting with side b using cos c = cos a cos b cos b = cos c cos a. So cos 78.42 cos b = cos b = 0.57914 cos 69.72 b = 54.6097. We find B using So Lastly we find A using So cos B = tan a cot c. cos B = tan 69.72 cot 78.42 cos B = 0.5545297899 B = 56.3220. cos A = sin B cos a. cos A = sin 56.32 cos 69.72 cos A = 0.28843 A = 73.24. 13 e) Given A = 52.4 and B = 122.27, we begin with So Next we find b using So Lastly we find length a using So cos c = cot B cot A. cos c = cot 122.27 cot 52.4 cos c = 0.48628 cos b = c = 119.0961. cos b = cos B sin A. cos 122.27 sin 52.4 cos b = 0.67388 b = 132.3674. cos a = cos A sin B. cos 52.4 cos a = cos a = 0.74160 sin 122.27 a = 43.8180. [AP ], [T T ]
14 5. CHAPTER 5 SOLUTIONS 6) b) A spherical triangle s angles must sum to > 180. Since C = 90 and A = B, both A and B must be > 45. Since a = b, cos c = cos a cos b is always non-negative. Therefore, by the spherical Pythagorean Theorem, cos c is always non-negative c 90. [T T ], [AP ] 9) b) Using cos c = cos a cos b and cos B = cos b sin A, we solve for cos b in the second, and substitute it into the first and then simplify to get sin A cos c = cos a cos B. [T T ], [AP ] 10) a) We first rearrange our spherical identity as follows: Since sin b sin c b c as b, c 0, we have sin B = sin b sin c. sin B = b. [AP ], [T T ] c
6 Chapter 6 Solutions 2) Given the angles A = 100, B = 69, C = 84 we can use the law of cosines for angles to determine side c. Solving for c we find ( ) cos 84 c = cos 1 + cos 100 cos 69 = 87.36. sin 100 sin 69 Rearranging the variables in the law of cosines for angles, we determine a and b: ( ) cos 69 b = cos 1 + cos 100 cos 84 = 69.67. sin 100 sin 84 ( ) cos 100 a = cos 1 + cos 69 cos 84 = 98.43. sin 69 sin 84 We determine each arc length by dividing them by 360 and multiplying by the circumference of the circle. Since this applies to all three sides, we can instead use it on the sum of the sides 87.43 + 69.67 + 98.43 = 255.46, thus the perimeter is 255.46 360 2 π 10 in = 44.58 in. [T T ], [AP ] 15
16 6. CHAPTER 6 SOLUTIONS 4) Figure 6.1: The initial heading. We will use ABC with B at our initial point, C at the south pole, and A at our final position. In Figure 6.1 the arc containing BC is E79 and the arc containing AC is E52. Since we are crossing the meridian lines, we know C = 27. Because c = 2060 nautical miles, then c = 34.33. To find b, the final latitude, we can use the sine law b = sin 1 ( sin 34.33 sin 50 sin 27 ) = 72.10 Because this is the distance from the pole to A, our latitude must be 90 b = 17.90 S. Our final bearing is A, and we can use Napier s second analogy to determine ( A = 2 tan 1 sin ( 1 2 (72.1 34.33 ) ) ) sin ( 1 2 (72.1 + 34.33 ) ) tan ( 1 2 (50 27 ) ) = 126.56, so our bearing is 180 A = 53.44. Finally we can use Napier s third analogy to find our initial latitude a: ( ( cos 1 a = 2 tan 1 2 (50 + 27 ) ) tan ( 1 2 (72.1 + 34.33 ) ) ) cos ( 1 2 (50 27 ) ) = 93.77. From this value our inital latitude can be determine using a 90 = 3.77 N. [AP],[TT]
13) Construct the ABC with A at Vancouver, B at Edmonton, and C at the north pole 17 Figure 6.2: Vancouver to Edmonton. where C = 9.6, the difference between the meridians. We are given the latitude for the two cities: Vancouver= 49.3 N and Edmonton= 53.6 N, so we find a = 36.4 and b = 40.7. Using Napier s first and second analogies we can construct a system of equations for A + B and A B which will allow us to find both angles. ( A B = 2 tan 1 sin ( 1 2 (36.4 40.7 ) ) ) sin ( 1 2 (36.4 + 40.7 ) ) tan ( 1 2 (9.6 ) ) = 71.27, A + B = 2 tan 1 ( cos ( 1 2 (36.4 40.7 ) ) ) cos ( 1 2 (36.4 + 40.7 ) ) tan ( 1 2 (9.6 ) ) = 172.48. Combining (1) and (2): A = 50.61 and B = 121.87. We now use the law of sines to determine the distance between the cities ( ) sin 9.6 c = sin 1 sin 36.4 = 7.35. sin 50.61 To find the distance, knowing the radius of the earth is 6371 km, 7.35 2 6371 km π = 817.28 km. [T T ], [AP ] 360
18 6. CHAPTER 6 SOLUTIONS 14) a) We begin with two cosine laws, one for A cos A = cos B cos C + sin B sin C cos a and another for B cos B = cos A cos C + sin A sin C cos b. We add them together to get cos A + cos B = cos B cos C cos A cos C + sin B sin C cos a + sin A sin C cos b. We simplify: and find cos A + cos B = cos C(cos B + cos A) + sin C(sin B cos a + sin A cos b) (cos A + cos B)(1 + cos C) = sin C(sin B cos a + sin A cos b). (1) From the Law of Sines we have sin B = m sin b and sin A = m sin a. Substitute them into (1): b) Taking (cos A + cos B)(1 + cos C) = sin C(m sin b cos a + m sin a cos b). From the sine addition law, we get (cos A + cos B)(1 + cos C) = m sin C sin(a + b). [AP ], [T T ] sin A + sin B sin a + sin b = m and dividing it by our result from a), we find Simplify: sin A + sin B sin a + sin b 1 (cos A + cos B)(1 + cos C) = m m sin C sin(a + b). sin A + sin B sin a + sin b = cos A + cos B sin(a + b) 1 + cos C sin C. c) Converting sin(a + b) into sin ( 2 ( )) ( ( a+b 2 and sin C into sin 2 C )) 2, we find ( ) 2 sin A+B cos A 2 a+b a b 2 sin cos 2 2 2 2 2 cos 2 C 2 = 2 cos A+B cos A B 2 sin C cos ( ). C 2 2 2 2 2 sin a+b a+b cos 2 2 d) Simplifying gives tan A+B 2 cot C 2 = a b cos 2 cos a+b 2. [T T ], [AP ]