Notes on Spherical Triangles

Transcription

1 Notes on Spheril Tringles In order to undertke lultions on the elestil sphere, whether for the purposes of stronomy, nvigtion or designing sundils, some understnding of spheril tringles is essentil. The ommonly used formule for spheril tringles re most redily derived using vetor nottion. Aordingly, these notes egin with disussion of vetors. Vetor Identities Three vetor identities re prtiulrly useful. These will e exploited when deriving the formule used for spheril tringles. Identity I Given three ritrry vetors x, y nd z: (x y).z = (y z).x = (z x).y Proof: eh expression is the volume of the sme prllelepiped. Identity II Given three ritrry vetors x, y nd z: x (y z) = (x.z)y (x.y)z Proof: tke three mutully perpendiulr unit vetors i, j nd k rrnged so tht y nd j re ligned nd tht z is in the plne defined y j nd k. The third vetor, x, is ritrrily ligned. The following figure shows the sheme: z y k j x i Fig. 1 Three Aritrry Vetors 1

2 Given the rrngement shown in Fig. 1, slr vlues x 1, x 2, x 3, y 2, z 2 nd z 3 n e found suh tht: x = x 1 i + x 2 j + x 3 k y = z = Noting tht j j = 0 nd tht j k = i: y 2 j z 2 j + z 3 k y z = y 2 z 3 i Hene: x (y z) = x 2 y 2 z 3 k + x 3 y 2 z 3 j = x 3 y 2 z 3 j x 2 y 2 ( z 2 j + z 2 j + z 3 k) = x 3 z 3 y 2 j x 2 y 2 ( z 2 j + z) = (x 2 z 2 + x 3 z 3 )y 2 j x 2 y 2 z = (x.z)y (x.y)z Identity III Given four ritrry vetors p, q, r nd s: (p q).(r s) = (p.r)(q.s) (p.s)(q.r) Proof: (p q).(r s) = ( q (r s) ).p...y Identity I Then, pplying Identity II to the right-hnd side: (p q).(r s) = ( (q.s)r (q.r)s ).p = (p.r)(q.s) (p.s)(q.r) 2

3 Spheril Tringles A spheril tringle is region on the surfe of sphere ounded y the rs of three gret irles. Without loss of generlity, the sphere n e deemed to hve unit rdius. A typil spheril tringle is shown in Fig. 2: A B C O Fig. 2 A Spheril Tringle The entre of the sphere is shown s point O. The lengths of the three rs ounding the tringle re shown s, nd. These lengths re mesured s ngles sutended y the rs t the entre of the sphere. The three vlues A, B nd C re the ngles t the verties of the spheril tringle. Conventionlly the side whose length is is rrnged opposite the vertex whose ngle is A nd so on. Three vetors,, nd, re lso shown in the figure. These re the outwrd vetors from the entre of the sphere to the three verties. Clerly the length of eh of these vetors is the rdius of the sphere nd, ssuming tht the sphere hs unit rdius,, nd re unit vetors. The First Cosine Rule The first of the formule for spheril tringles is: os = os os + sin sin os A To derive this, use Identity III to give: ( ).( ) = (.)(.) (.)(.) (1) 3

4 Now onsider the following view of vertex A looking diretly towrds the entre of the sphere long : A A Fig. 3 Close-up of Vertex A looking long Note tht is perpendiulr to the plne defined y nd nd hene to r whih is in this plne. Likewise is perpendiulr to r. Sine the ngle etween r nd r is A, the ngle etween nd is lso A. Given tht is the ngle etween nd, the mgnitude of is sin. Likewise the mgnitude of is sin. Sine the ngle etween nd is A the left-hnd side of (1) is: ( ).( ) = sin os A sin The right-hnd side of (1) is trivilly: (.)(.) (.)(.) = 1.os os os Equting the left- nd right-hnd sides leds diretly to the first osine rule. A speil se is of pssing interest. The following figure shows spheril tringle in whih vertex A is pprohing 180 : A Fig. 4 Approhing Degenerte Cse When A=180, os A = 1 nd the first osine rule gives: os = os os sin sin = os( + ) This leds to the ovious result tht, in the limit, =+. 4

5 The Seond Cosine Rule The seond of the formule for spheril tringles is: os A = os B os C sinb sinc os To derive this, set up three unit vetors x, y nd z prllel to, nd respetively, s illustrted in the following figure: A B z C x O y Fig. 5 A Spheril Tringle From Identity III: (z x).(x y) = (z.x)(x.y) (z.y)(x.x) (2) Now trnslte vetors x nd z to vertex B nd onsider the following view of tht vertex looking diretly towrds the entre of the sphere long : z x B B Fig. 6 Close-up of Vertex B looking long Note tht x is prllel to whih is perpendiulr to the plne whih ontins r, nd hene x is perpendiulr to r. Likewise z is perpendiulr to r. Sine the ngle 5

6 etween r nd r is B, it is ler tht the ngle etween x nd z is 180 B. This ngle is shown s B in Fig. 6. Given tht z nd x re unit vetors, the mgnitude of z x is sinb whih is the sme s sin B. The diretion of z x is towrds the entre of the sphere long. Sine is unit vetor whose diretion is outwrds from the entre of the sphere, z x = sin B. Likewise x y = sinc. Aordingly, the left-hnd side of (2) is: (z x).(x y) = ( sin B).( sinc) = sin B os sinc Given tht z nd x re unit vetors, the mgnitude of z.x is os B whih is the sme s os B. Likewise x.y = os C nd z.y = os A. Aordingly the right-hnd side of (2) is: (z.x)(x.y) (z.y)(x.x) = ( os B)( os C) ( os A).1 = os B os C + os A Equting the left- nd right-hnd sides leds diretly to the seond osine rule. The Sine Rule The third of the formule for spheril tringles is: To derive this, use Identity II to give: sina sin = sinb sin = sinc sin ( ) ( ) = ( ( ). ) ( ( ). ) (3) With referene to Fig. 3, the mgnitudes of nd hve lredy een shown to e sin nd sin nd the ngle etween them is A. Aordingly the left-hnd side of (3) is: ( ) ( ) = sin sin A sin The diretion of this vetor n e verified to e y inspetion of Fig. 3, The seond term on the right-hnd side inorportes ( ). whih is zero (vetors nd re orthogonl so their dot produt is zero). Equting the left-hnd side to the remining term on the right-hnd side gives: sin sina sin = ( ). nd similrly sin sin B sin = ( ). (4) By Identity I, the two right-hnd sides of (4) re identil, so: sin sina sin = sin sinb sin sina sin = sin sinb This leds diretly to the first prt of the sine rule nd, y rottion of identifiers, to the seond prt. Note tht for smll tringles sin, sin nd sin nd the sine rule for spheril tringles is redily seen to e equivlent to the ordinry sine rule for plne tringles. 6

7 The First Tngent Rule The fourth of the formule for spheril tringles is: tn = tnasin sinc + tnaos os C The sides nd verties re lelled s efore with the unknown nd A, nd C known. A C Fig. 7 A Spheril Tringle To derive the first tngent rule, use the sine rule: Noting tht ot 2 θ = ose 2 θ 1: sin = sina sin sin B ot 2 = sin 2 B sin 2 A sin 2 1 So: By the seond osine rule: ot 2 = 1 os2 B sin 2 A sin 2 sin 2 A sin 2 (5) os B = os C os A sinc sin A os Use this expression for os B to rewrite the top line of the right-hnd side of (5): 1 (os 2 A os 2 C 2os A os C sina sinc os + sin 2 A sin 2 C os 2 ) sin 2 A sin 2 Reple the leding 1 y os 2 A + sin 2 A: = (os 2 A + sin 2 A) os 2 A os 2 C + 2os A os C sina sinc os sin 2 A sin 2 sin 2 A sin 2 C os 2 7

8 Group together terms whih ontin os 2 A nd terms whih ontin sin 2 A: = os 2 A(1 os 2 C) + 2os A os C sina sinc os + sin 2 A(1 sin 2 sin 2 C os 2 ) Simplify in stges: = os 2 A sin 2 C + 2os A os C sina sinc os + sin 2 A(os 2 sin 2 C os 2 ) = os 2 A sin 2 C + 2os A os C sina sinc os + sin 2 A(1 sin 2 C) os 2 = os 2 A sin 2 C + 2os A os C sina sinc os + sin 2 A os 2 C os 2 = (os A sin C + sina os C os ) 2 This is the top line of the right-hnd side of (5). Aordingly: Hene: ot = tn = = os A sinc + sin A os C os sina sin sina sin os A sinc + sina os os C tna sin sinc + tna os os C (6) (7) Speil Cses: If C=90, the first tngent rule redues to: tn = tna sin If, further, the tringle is smll, tn nd sin whih leds to: =.tn A This orresponds to the elementry use of the tngent funtion with plne right-ngled tringle. Note tht lthough (7) is the version of the first tngent rule whih ws originlly presented, it is sfer to use (6) if there is ny dnger of A pprohing 90 (when tna diverges). If A=90, (6) redues hrmlessly to: tn = sin os C os = tn os C If, further, the tringle is smll, tn nd tn whih leds to:.os C = This orresponds to the elementry use of the osine funtion with plne right-ngled tringle. 8

9 The Seond Tngent Rule The fifth of the formule for spheril tringles is: tn A = tnsin B sin tnos B os The sides nd verties re lelled s efore with A the unknown nd, B nd known. B A Fig. 8 A Spheril Tringle To derive the seond tngent rule, use the sine rule: Noting tht ot 2 θ = ose 2 θ 1: sina = sin sin B sin ot 2 A = sin 2 sin 2 sin 2 B 1 So: By the first osine rule: ot 2 A = 1 os2 sin 2 sin 2 B sin 2 sin 2 B os = os os + sin sin os B (8) Use this expression for os to rewrite the top line of the right-hnd side of (8): 1 (os 2 os 2 + 2os os sin sin os B + sin 2 sin 2 os 2 B) sin 2 sin 2 B Reple the leding 1 y os 2 + sin 2 : = (os 2 + sin 2 ) os 2 os 2 2os os sin sin os B sin 2 sin 2 B sin 2 sin 2 os 2 B 9

10 Group together terms whih ontin os 2 nd terms whih ontin sin 2 : = os 2 (1 os 2 ) 2os os sin sin os B + sin 2 (1 sin 2 B sin 2 os 2 B) Simplify in stges: = os 2 sin 2 2os os sin sin os B + sin 2 (os 2 B sin 2 os 2 B) = os 2 sin 2 2os os sin sin os B + sin 2 (1 sin 2 ) os 2 B = os 2 sin 2 2os os sin sin os B + sin 2 os 2 os 2 B = (os sin sin os os B) 2 This is the top line of the right-hnd side of (8). Aordingly: ot A = os sin sin os os B sin sin B Hene: tna = = sin sinb os sin sin os B os tn sin B sin tn os B os (9) (10) Speil Cses: If B=90, the seond tngent rule redues to: tna = tn sin If, further, the tringle is smll, tn nd sin whih leds to: tna = This orresponds to the elementry use of the tngent funtion with plne right-ngled tringle. As with the first tngent rule, lthough (10) is the version whih ws originlly presented, it is sfer to use (9) if there is ny dnger of pprohing 90 (when tn diverges). Frnk H. King 30 June