Spheril Geometry This is n rtile from my home pge: www.olewitthnsen.dk Ole Witt-Hnsen nov. 6
Contents. Geometry on sphere.... Spheril tringles...3. Polr tringles...4 3. The right-ngle spheril tringle...6 4. Clultion of sides nd ngles in the right-ngle spheril tringle...8 5. The generl spheril tringle. Cosine- nd sine reltions... 5. The osine reltions... 5. The sine reltions... 6. The Are (the surfe) of spheril tringle...3 7. Exmples nd exerises to the generl spheril tringle....4 7.5 Solving the homogeneous eqution in os x nd sin x...7 8. The plne geometry s limiting se of the spheril geometry...8 9. Exerises...
Spheril Geometry. Geometry on sphere It is fundmentl ft in Eulidin geometry tht the shortest pth between to points lies on t stright line between the two points. In n rbitrry two dimensionl surfe, things beome more omplex. The prt of mthemtis tht trets this re is lled differentil geometry. It is, however, fr more omplited thn ordinry geometry in plne. If the surfe is given by prmeter representtion P( u, u) ( x( u, u), y( u, u), z( u, u)), then using the theory of differentil geometry, one n derive (rther omplited) differentil eqution, where the solution is prmeter representtion of the shortest pth between two points on the surfe. Suh urve is lled geodeti. As geo mens erth in Greek, the onept of geodeti refers to the shortest pth between two points on the surfe of the erth. If one rnks the two dimensionl surfes fter inresing omplexity, the sphere omes in s number two fter the plne. Figur () Figur () If you ut sphere with plne, the interseting urve is irle. If the ut goes through the entre of the sphere the intersetion is gret irle, tht is, hving dimeter equl to the dimeter of the sphere. Both ses re shown in the figures bove. In the differentil geometry, one my show, (but it is fr from simple), tht the shortest pth between two points on sphere is prt of the gret irle through the two points. In the figure to the right, the two gret irles re e.g. the shortest distne between A nd B, nd B nd C. In the plne most geometril figures onsists of stright line segments, nd sine the gret irles, represents the stright lines, (the geodetis), we shll only be onerned with the geometry of figures onsisting of gret irle segments. A gret irle is uniquely determined by to points on the sphere A nd B, whih do not lie dimetrilly opposite. www.olewitthnsen.dk Differentil Geometry
Spheril Geometry 3 The intersetions between two different gret irles re two dimetrilly opposite points, nd the line onneting them is dimeter, whih is lso the intersetion line between the plnes tht genertes the two gret irles. The ngle between two gret irles is defined (it is) the ngle between the two orresponding plnes. From the figure (), it is obvious tht the r A B = 8 - AB. The dimeter (the xis) (tht is perpendiulr to plne tht belongs to gret irle) intersets the sphere in two dimetrilly opposite points A nd A, whih re nmed the poles belonging to the gret irle. When one of two gret irles intersets the other in its poles, the two gret irles re perpendiulr to eh other, s illustrted on the figure below. Figur (3) Figur (4). Spheril tringles A spheril tringle is prt of the sphere tht is onfined by the rs of three gret irles, nd they re lled the sides of the spheril tringle. The sides re mesured in degrees, or lterntively in rdins. The length, of side on sphere with rdius R is found by multiplying its rdin number α by R. = αr. The lbels, b, for the sides in spheril tringle, my denote the degrees, the rdins or the length of the sides dependent on the ontext. The intersetion ngles between the gret irles tht form the spheril tringle re the ngles of the tringle. The ngles lying opposite to the sides, b, re denoted A, B, C, s is the se in plne geometry. If lines re drwn form the entre of the sphere to the three verties of the tringle, it will form tringulr orner. Sides nd ngles in the tringulr orner re in pirs equl to the sides nd ngle in the spheril tringle, s depited in figure (4) bove.
Spheril Geometry 4 The onepts of isoseles, height, biseting lines nd bisetor norml re the sme s in the geometry of the plne. Figur (5) A right-ngled tringle is spheril tringle, hving (t lest) one ngle equl to 9. A spheril tringle might hve one, two or three right-ngles. Cthetus nd hypotenuse hve the sme mening s in plne right-ngled tringle, (if the tringle hs only one right-ngle). The lune (hlf moon) ABA C shown in figure (5) is lled spheril double edge. It hs only one ngle, whih is the ngle between the two plnes tht form the double edge. As we shll see the sum of the three ngles in spheril tringle re lwys greter thn 8. In figure (5) is shown tringle hving two right-ngles. If lso A = 9, the tringle hs three right-ngles, whih is possible. However, s spheril ngle nnot exeed 8, the sum of the ngles is lwys less thn 54.. Polr tringles By the polr tringle to spheril tringle, we understnd the tringles reted by the poles of the gret irles tht the sides of the given tringle re prt of. The ngles in the polr tringle re often denoted with n index. So A, B, C, re the poles belonging to the sides, b,. Figur (6) Figur(7) In the figure to the left, the pole A belonging to the side, is onstruted. In the figure to the right the whole polr tringle A B C is shown. But obviously it is diffiult to visulize sptil figure, even when it is drwn in perspetive, nd espeilly, when it is not drwn professionlly. For exmple, in figure (7) both B nd C lie on the bk side of the sphere, while A lies in the pln of the pper. A simple but rther non trnsprent theorem sttes: The polr tringle to given spheril tringles polr tringle is the tringle itself.
Spheril Geometry 5 If we onsider the ft tht when one of two gret irles psses through the poles of the other gret irle, then the two gret irles re perpendiulr to eh other, s is depited in figure (3), we reson s follows Let the given spheril tringle be A, B, C nd its polr tringle A B C. We seek the pole for the side = B C. As B C intersets pole for eh of the gret irles AB nd AC, we n onlude tht AB nd AC pss through the poles for B C. Consequently one of the intersetion points must be A. Furthermore sine AA < 9, A lies on the sme side of B C s A does, then A must be the pole to be used when onstruting the polr tringle to the spheril tringle A B C. A similr resoning my be used to onstrut the poles C nd B to A B nd A C. This is supplemented by somewht surprising theorem. The sides (, b, ) nd ngles (A, B, C ) in the polr tringle re equl to the omplementry ngles of the ngles (A, B, C) nd sides (, b, ) in the originl spheril tringle respetively. Sine B nd C lie on the norml to the plnes orresponding to the sides AC nd AB, nd s the ngle between the norml to two plnes is 8 - v, where v is the ngle between the plnes themselves, nd sine A is the ngle between the plnes orresponding to the sides AC nd AB, then = B C = 8 A. A quite similr rgument n be pplied for the other two sides of the polr tringle. Applying the bove theorem, tht the polr tringle to polr tringle is the tringle itself, we find AB= = 8 C or C = 8, nd similrly for the two other ngles. Thus the theorem is proved. Applying the bove theorem bout the sides nd the ngles in the polr tringle to spheril tringle, we n prove tht the sum of the three ngles in spheril tringle is lwys bigger thn 8 nd less thn 54. We ssume tht we hve onstruted the polr tringle to spheril tringle ABC. Aording to the bove theorem the sides in the polr tringle re 8 A, 8 B, 8 C. sine the sum of the three sides must be less thn 36, the following inequlity is vlid: (.) 8 A+ 8 B+ 8 C < 36 A + B + C > 8 Furthermore the sum of the sides in the polr tringle must be greter thn (.) 8 A+ 8 B+ 8 C > A + B + C < 54 The differene E = A + B + C - 8 is lled the spheril exess.
Spheril Geometry 6 3. The right-ngle spheril tringle Figur(8) Let the tringle be ABC, where C = 9. O is the entre of the sphere, nd we ssume tht the rdius is, so tht the line segments OA, OB nd OC ll hve the length. D is the projetion of B on OC, nd E is the projetion of B on OA. We ssume tht the two thetus nd b re ute, so tht D lies between O nd C, nd E lies between O nd A. DE is perpendiulr to OA, sine ED is the projetion of BE on the plne OAC. As BE nd ED both re perpendiulr to OA, then BED is n ngle between the plnes AOB nd AOC, nd thus BED = A. At the sme time the hypotenuse = BOE. We re frequently going to pply the formuls for the plne right-ngle tringle, so we write them below, both s formuls nd s sttements. For ny plne right-ngle tringle, the following holds: (3.) sin A os A b tn A b Sinus to n ngle is equl to the opposite thetus divided by the hypotenuse. Cosine to n ngle is equl to the djent thetus divided by the hypotenuse. Tngent to n ngle is equl to the opposite thetus divided by the djent thetus For the tringulr orner in figure (8) is thus seen: OD OB os os nd OE OD osb os osb From the right-ngle tringle OEB we further get: OE OB os os, thus (3.) os os osb From the right-ngle tringle OEB we get: OE OB os nd from ΔBED we find: BD sin A, but BD = sin nd BE =, so BE (3.3) sin sin A
Spheril Geometry 7 Similrly to this we hve: (3.4) ED sin b sin B, so BE sin b sin B Also we find from BDE tht DE os A, nd sine DE OD sin b os sin b, we get BE (3.5) os sin b os A orrespondingly osbsin os B We hve derived the formuls bove under the ondition tht both ngles A nd B re ute. If C= 9, nd one thetus e.g. = 9, then A = 9, whih lso ppers from the formuls. Figur (9) Figur () A spheril tringle n be onsidered s resent ( hlf moon) onneting two poles C nd C where the ngles C = C, interseted by gret irle, thereby reting two spheril tringles ACB nd AC B. They re lled neighbouring tringles. C AB 8 CAB, nd C BA 8 CBA. Every spheril tringle hs three neighbouring tringles, one for eh side. If both thesus re obtuse ngles (mening greter thn 9 ) wht is shown in figure (9) we fous on the neighbour tringle A B C, with sides, b,. The points A =A, B =B, C nd C re opposite poles. = 8 -, b = 8 - b, =. In the tringle A B C both thetus re ute, so we my pply the formuls (3.) - (3.5). sin sin(8 ) sin A sin(8 A) sin A sin(8 ) sin
Spheril Geometry 8 Sine sin(8 v) = sinv, we rrive t the sme formuls s before os sin b os(8 )sin(8 b) os A os(8 A) os A sin(8 ) os os os(8 )os(8 b) os osb os sin b Sine os(8 v) = - osv, we lso find the sme formuls s for the tringle with ute thetus. If > 9 nd b < 9, we my fous on the neighbour tringle A B C, to the side b, (the figure to the right), whih hs n ngle 9 nd ute ngle, nd ording to the preedents, the following formuls re vlid. os os(8 ) os(8 )osb os os osb sin sin(8 ) sin A sin(8 A) sin A sin(8 ) sin os sin b os(8 )sin b os A os(8 A) os A sin(8 ) The formuls derived bove re therefore vlid for ny right-ngle spheril tringle. 4. Clultion of sides nd ngles in the right-ngle spheril tringle. From the formuls (3.) - (3.5) we my further derive some formuls involving tn. os sin b From: os os osb os os osb nd os sin b os sin b tn b os A osbsin tn tn b (4.) os A tn sin os sin b By dividing sin A with os A, we get, tn (4.) tn A sin b And its nlogous formul tn b tn B sin tn A sin os sin b tn sin b By multiplition of (4.) nd (4.) we get: tn tn b sin sin b tn Atn B sin b sin os sin b osbsin os osb os
Spheril Geometry 9 (4.3) tn Atn B os os tn A tn B Below we hve olleted ll the formuls belonging the right-ngle spheril tringle. (4.4) os os osb (4.5) (4.6) (4.7) sin sin A nd os sin b os A nd tn b os A tn nd sin b sin B osbsin os B tn os B tn (4.8) tn Atn B os os tn A tn B 4.9 Exmple: = 35, b = 6. os = os os b = os 35 os 6 =.495 => = 66.8 sin sin 35 sin A.687 sin 65.8 sin b sin 6 sin B,9493 sin 65.8 A 38.96 B 7.68 4. Exmple: = 4-5, =.56 os os os osb osb.4 os sin sin A.637 A 4.9 sin b sin B 9875 B 8.93 b 3.89 4. Exmple: = 36.7, A = 5.83
Spheril Geometry sin sin sin A.77,64 5.93 sin A os os os osb osb.786 b 38.8 os sin b sin B.796 B 5.77 4. Exmple: A = 43.8, B = 8.59 os.76 tn A tn B sin sin Asin.676 sin b sin Bsin.97 79.86 4.45 76. 5. The generl spheril tringle. Cosine- nd sine reltions Figur() Figur() 5. The osine reltions We shll then onsider the generl spheril tringle s shown from figure (). We drw the height from one of the ngles e.g. from B. The low end of the height is D. Firstly we ssume tht D lies between A nd C. AD is denoted x, so tht DC beomes b x. We then pply formul (4.4) on ΔABD nd ΔBDC. os os x os h nd os os hos( b x) Then we shll use the ddition formuls for osine: eqution bove. os( u v)osu os v sin u sin v in the lst
Spheril Geometry os os h os( b x) os hosbos x os hsin bsin x And with help from (4.4) os os x os h the bove eqution n be written: os osbos os hsin bsin x Using (4.6) we get os sin b os A on ΔABD os hsin x os A os hsin x os Asin We thus end with the osine reltion, vlid for the generl spheril tringle. (5.) os osbos sin bsin os A (5.) hs (s is the se for the osine reltion for the plne tringle) two nlogous expressions, whih omes bout, by permuting the letters. (5.) osb os os sin sin os B (5.3) os os osb sin sin bosc The osine reltions re suitble to find the ngles in spheril tringle, when the three sides re given. If the low point of the height lies outside AC, s shown in figure () the lultions re lmost identil to tht of the former se. From ΔABD it follows s before, from (4.4) : os os hos x, nd from ΔBDC we get: os os h os( x b) os h(os x osb sin xsin b) os hos x osb os hsin xsin b When we insert the expression for os, we get: os osbos os hsin xsin b As we did before, using (4.6) os sin b os A on ΔABD, we find os( 8 A)sin os hsin x When inserted in the expression for os, we end with the sme expression for the osine reltion s previously. os osbos sin bsin os A
Spheril Geometry In ontrst to wht holds true for the plne tringle, spheril tringle is omplete determined by its three ngles. To try to solve the three osine reltions to determine, b, is not lgebrilly pltble, lthough there re three independent lgebri equtions with three unknowns. Insted of trying one should rell the theorem mentioned bove: In polr tringle, whih is formed by the poles of the three gret irles, (of whih the three sides,b, belong to), the ngles A, B, C of the polr tringle re omplementry ngles to the sides,b, in originl spheril tringle ABC, nd the sides,b, in the polr tringle re omplementry ngles to A,B,C in the originl spheril tringle. e.g. = 8 - A Writing the three osine reltions belonging to of the sides the polr tringle, we get three equtions to determine the sides in the originl tringle. (5.4) os osb os sin b os A Whih ording to wht is stted bove is equl to os( 8 A) os(8 B)os(8 C) sin(8 B)sin(8 C) os(8 ) os A ( os B)( osc) sin Bsin C( os ) os A os BosC sin Bsin C os From whih one n determine. Formuls for the two sides b nd n be obtined, by permuting the letters ylilly. 5. The sine reltions sin If we pply the formul (4.5): sin A to the two right-ngle tringles ΔABD nd ΔBDC in figure (), then it holds true, whether or not the height hs it low end between B nd C, tht: sin h sin A nd sin h sinc sin sin h sin Asin nd sin h sinc sin From whih it follows: sin Asin sin C sin Resulting in the sine reltions for ny spheril tringle. (5.5) sin sin A sin B sin b sin C Where the term in the middle is the result of permuting the letters.
Spheril Geometry 3 You should however ontemplte tht, when using the sine reltions to find n ngle, the eqution sin v = x hs two solution v nd 8 v. Whih one belongs to the tringle or the neighbouring tringle my be resolved by drwing test tringle, or using the osine reltions. 6. The Are (the surfe) of spheril tringle Figur(3) The surfe of sphere with rdius R is T = 4πR. Figure (3) shows spheril tringle ABC. We hve lso drwn the neighbour tringle to the side, whih is A BC, nd the neighbour tringle to the side b: AB C, nd to the side : ABC. The points A nd A, B nd B, C nd C lie dimetril opposite, nd ny tringle nd its neighbouring tringle form double edge, ( hlf moon) with n ngle tht is equl to the ngle t the top. For exmple, the two tringles ABC nd A BC form double edge with the ngle A = A. The re of double edge seprted by n ngle must be /36 times the re of the sphere. Consequently the re of double edge seprted by n ngle v is (6.) T v v T 36 v 4R 36 From this follows, writing: T(ABC) for the re of ΔABC. A T( ABC) T( A BC) T 36 B (6.) T( ABC) T( ABC ) T 36 C T( ABC) T( ABC) T 36 And by dding the three equtions: A B C 36 (6.3) T ( ABC) T ( ABC) T( A BC) T( AB C) T ( ABC ) T C AB lies symmetrilly with CA B, with respet to entre of the sphere, so the two tringles hve the sme re. If you look t figure (3) you my onvine yourself tht the four tringles in the prenthesis in (6.3) together form hlf sphere, nd the terms therefore re equl to T. Inserting this in (6.3) gives:
Spheril Geometry 4 (6.4) A B C 8 A B C T ( ABC) T T T ( ABC) T T 36 36 36 A B C 8 T ( ABC) T 36 E E T( ABC) T T ( ABC) R 7 8 This is the generl expression for the re of spheril tringle, where T = 4πR (6.5) E A B C 8 is lled the spheril exess. It is remrkble ft tht the re of spheril tringle neither depends on the size of the sides nor the size of the ngles, but only on the spheril exess. 7. Exmples nd exerises to the generl spheril tringle. Exmple 7. Given the three sides: = 8, b =, = 65. It is lwys good ide to drw plne test tringle, whih does not need to be very urte. The three ngles A, B, C n be determined using the osine reltion. os osbos sinbsin os A os osbos os A sinbsin os8 os os 65 os A.736 sinsin 65 And in the sme mnner: A.93 osb osos os B sin sin.4654 B 7.74 os ososb osc.54 sin sinb Exmple 7. C 57.5 Given the three ngles: A = 8, B =, C = 65. One might be inlined to solve the three osine reltions to obtin, b,, but it will led nowhere. Insted one should use the polr tringle A, B, C, s we hve shown previously the ngles nd sides re omplementry (8 v) to the sides nd ngles in the originl tringle.
Spheril Geometry 5 =8 - A, b = 8 B, =8 - C nd subsequently determine A = 8 -, B =8 - b, C = 8 -. With the given vlues for the ngles, whih re the sme s the sides in exmple 7., we only need to write: 8.93 58.7, b 8 7.74 6. 6, 8 57.5. 49. Exmple 7.3 To void tht this looks like pure mthemtil mgi, we repet exmple 7., but with fresh vlues for the ngles. A = 57, B = 75, C =. We then proeed by writing down the osine reltions for the polr tringle, where =8 - A, b =8 - B, = 8 - C, nd then determine A, B, C. Finlly we determine = 8 - A, b = 8 - B, = 8 - C. os osb os sinb os A os(8 A) os(8 B)os(8 C) sin(8 B)sin(8 C)os(8 ) os A osbosc sin BsinC os os A os B osc os sin B sin C.553 58.3 And by permuting the letters: os B os AosC osb sin Asin C.989 b 78.53 osc os Aos B os.43 sin Asin B Exmple 7.4 Given n ngle nd the two djent sides: 9.3 = 8, b = 43, C = 59. The side n diretly be determined by the osine reltion: os os osb sin sin bosc.48 9.7
Spheril Geometry 6 B nd C n then in priniple be found by the sine reltions: sin sin A sin B sinb sinc sin B sin C sin bsin C sin B.5 B 3. B 66.99 sin b sin A sin C sin sin C sin A.86 A.84 A 59.6 sin The problem is of ourse tht we get two solutions for eh ngle, nd tht we relly hve no wy to deide whih ngle belongs to the given tringle, nd whih one to the neighbour tringle to the side. In plne trigonometry the sme tringle hs two solutions, but in spheril geometry there is only one solution, the other solution belongs to the neighbouring tringle. In some ses it is ler whih ngle to hoose, if you drw plne test tringle. To irumvent this problem, without resorting to plne test tringle we turn insted to the osine reltion to determine the ngles A nd B. os osb os os8 os43os6,7 os A.9345 sin bsin sin43sin6.7 A 59.6 osb os os os B sin sin.9774 B 66.98 Exmple 7.5 In the preeding exmples, we hve been ble to find unknown sides nd ngles lmost in the sme mnner s in the plne geometry. But in spheril geometry problems rise, if we onsider the two ses:. Given n ngle, n djent nd the opposite side, e.g. A,, b.. Given side, n djent nd n opposite ngle, e.g. A, B,. The two ses re in ft the sme, sine B n be lulted from the sine reltions in the first se, nd b n be lulted from the sine reltions in the ltter. But in both ses we re left with A, B,, b, but re missing nd C. The problem rises of ourse from the ft, tht we n not find C = 8 (A+B), s we do in the plne geometry. So we re left with the osine reltion. It is orret tht we my write two osine reltions, whih only hve the side s unknown, but the problem is tht the unknown side ppers in the eqution both s os nd. Attempts to drw the height from C, nd the using the formuls for the right-ngle spheril tringle led nowhere. Alterntively we my write the osine reltion for os nd os b. (7.5.) os osbos sinbsin os A nd osb os os sin sin os B
Spheril Geometry 7 Then we solve both equtions for, nd put the results equl to eh other. os osbos nd sinbos A osb os os sin os B Whih gives n eqution with the unknown os, from whih n be determined. os osbos osb os os sinbos A sin os B One is determined, the ngle C my then be found from the osine reltion for. There is however nother possibility, sine both of the two osine reltions is so lled homogeneous eqution of first degree in osine nd sine. 7.5 Solving the homogeneous eqution in os x nd sin x The generl homogeneous eqution in os x nd sin x my be written: (7.5.3) os x bsin x To solve the eqution we divide by b. b os x b b sin x b Introduing the ngle y by: os y b nd sin y b b, It then follows (7.5.4) b tn y And the eqution beomes herefter os y os x sin y sin x b And it is rewritten with the help of the ddition formuls for os(x - y) os( x y) b The eqution hs solutions if:
Spheril Geometry 8 b. And the solution my be written x y os b 8. The plne geometry s limiting se of the spheril geometry The rdin number α for n r on gret irle hving rdius R is given by: R If the sides (the rs) in spheril tringle re very smll ompred to the irumferene, then the spheril tringle lmost ppers s plne. (We oneive the erth s flt). And therefore we should expet tht the trigonometri formuls for the spheril tringle in this limit re the sme s for the plne tringle. Tht this indeed the se, one n relize if we mke Tylor expnsion to the first signifint order of the trigonometri funtions. Resulting in: sin nd os The right-ngle spheril tringle We invent the nottion,,, for the rdin number of the sides, b, in the spheril tringle, nd t the sme time ssume tht: α <<, β <<, γ<<. The lengths of the sides then beome: = αr, b = βr, = γr. The formul (4.5) for the right-ngle spheril tringle then beomes: sin sin sin A sin R sin A R Whih, is seen to be identil to the formul for the plne right-ngle tringle. sin A By the sme token, when the formul (4.4) for the right-ngle spheril tringle os os osb is written with rdins, followed by Tylor expnsion to the seond order, nd keeping only terms up to the seond order.
Spheril Geometry 9 os os os ( )( ) R R R b We thus find tht the formul os os osb for the right-ngle spheril tringle is equivlent to Pythgors theorem for the right-ngle plne tringle. The generl spheril tringle: We shll first look t the sine reltions for the spheril tringle. sin sin A sin sin B b sin sin C When they re written with rdins sin A sin B sin C. sin sin sin We expnd the sine in the denomintor followed by multiply with R. sin A sin B sin C R R R sin A sin B sin C sin sin b Whih re seen to be the sine reltions for the plne right-ngle tringle. Next we turn to the osine reltions for the spheril tringle. os os osb sin sinbosc And we write it using rdins, followed by Tylor expnsion to the seond order. os os os sin sin os C ( )( ) osc osc osc R R R RR osc b b os C Whih, we reognize s the osine reltion for the plne tringle.
Spheril Geometry 9. Exerises. Oporto in Portugl nd New York City in USA lies lmost on the sme ltitude, s NYC = (4 45' n ; 74 ' w) nd Oporto = (4 45' n ; 8 4' w). (n: Northern ltitude, w: western longitude, nd 45 mens 45 r minutes) If one should sil or fly from Oporto to New York, it is most likely to trvel diretly west. ) Explin why stright west is not the shortest route, nd lulte the distne you my spre, by tking the shortest pth, nd present the result in nutil miles. To your informtion R erth = 637 km nd nuti mile (sm) = 854 m. Copenhgen lies t (55 4' n; 35' e) los Los Angeles lies t (34 ' n; 8 ' w). Clulte the spheril distne between CPH nd LA. On flight route goes vi Søndre Strømfjord in Greenlnd (66 ' n; 54 w ). Give n explntion why. (The plne does not lnd in Greenlnd to pik up pssengers) Clulte the spheril distnes Copenhgen - Søndre Strømfjord Los Angeles, nd ompre it to the spheril distne Copenhgen - Los Angeles. 3. The Bermud tringle is territory limited by Mimi: (5 49' n ; 8 6' w) - Puerto-Rio: ( '; 63 ') nd Bermud: (3 45' ; 65 ' ). The intension of this exerise is not to mke sttement onerning the mysteries in the Bermud tringle, but only to determine the ngles nd sides of the Bermud tringle, nd finlly give n indition of its re of the Bermud tringle in km. Solution to exerise. Eh time I hve given spheril geometry s finl ssignment in the Dnish yer high shool, I hve inluded exerise. Atully no student hs ever been ble to give the orret nswer (without help). So I hve hosen to revel the solution here. Although it is in ordne with spheril geometry, the onlusion is pt to use septiism mong non mthemtiins. To determine the length of, we mke use of the osine reltion Sine longitude nd ltitude re given in degrees nd r minutes, we initilly onvert it to deiml degrees. This is done by diving by 6 nd multiplying by, e.g. 4.45 = 4.75 nd 8.4 = 8.67. The rs = b = 9-4.75 = 49.5, nd C =74-8.67 = 64.33.
Spheril Geometry os os osb sin sinbosc os os 49.5os 49.5 sin 49.5sin 49.5os 64.33.6775 47,57 47.57 rd. 83 rd 8 The length is obtined by multiplying by the rdius R = 637 km of the erth. d =R =588 km = 588 sm 853 854 sm Whih is the distne one should trvel between Oporto nd New York long gret irle. If you lterntively trvel diret west, then you trvel by minor irle with rdius r =R sin, whih is r = 637 sin 49.5 km = 485 km The r tht you trverse is the r between the two ltitudes equl to C 64.33. 8 To determine the length d r you should multiply the r C with the rdius r. rd d r = 485.8 km = 547 km = 9 sm. The differene in nutil miles (sm) is therefore 9 853 = 69 sm = 8 km. The exmple onfirms in prtil mnner tht the shortest pth between to points on sphere is long gret irle.