Spherical Trigonometry

1 Spherical TrigonometryRob JohnsonWest Hills Institute of Mathematics1 IntroductionThe sides of a Spherical triangle are arcs of great circles. Agreat circle is the intersection of a sphere witha central plane, a plane through the center of that sphere. The angles of a Spherical triangle are measuredin the plane tangent to the sphere at the intersection of the sides forming the avoid conflict with the antipodal triangle, the triangle formed by the same great circles on the oppositeside of the sphere, the sides of a Spherical triangle will be restricted between 0 and radians. The angleswill also be restricted between 0 and radians, so that they remain derive the basic formulas pertaining to a Spherical triangle, we use plane Trigonometry on planes relatedto the Spherical triangle. For example, planes tangent to the sphere at one of the vertices of the triangle,and central planes containing one side of the specified otherwise, when projecting onto a plane tangent to the sphere, the projection will be fromthe center of the sphere.

2 Since each side of a Spherical triangle is contained in a central plane, the projectionof each side onto a tangent plane is a line. We will also assumethe radius of the sphere is 1. Thus, thelength of an arc of a great circle, is its 1: Central Plane of a Unit Sphere Containing the Side 1 One of the simplest theorems of Spherical Trigonometry to prove using plane Trigonometry is The SphericalLaw of (The Spherical Law of Cosines):Consider a Spherical triangle with sides , , and ,and angle opposite . To compute , we have the formulacos( ) =cos( )cos( ) +sin( )sin( )cos( )( )Proof:Project the triangle onto the plane tangent to the sphere at and compute the length of the projectionof in two different ways. First, using the plane Law of Cosines inthe plane tangent to the sphere at , wesee that the length of the projection of istan2( ) +tan2( ) 2tan( )tan( )cos( )( )Whereas if we use the plane Law of Cosines in the plane containing the great circle of , we get that thelength of the projection of issec2( ) +sec2( ) 2sec( )sec( )cos( )( )By applying Figure 1 to and , Figure 2 illustrates these two methods of computing the length of theprojection of onto the plane tangent at , that is, the red segment:Figure 2: Two ways to measure the red segmentSubtracting equation ( ) from equation ( ), we get that0 = 2 + 2tan( )tan( )cos( ) 2sec( )sec( )cos( )( )Solving forcos( ), gives The Spherical Law of Cosines:cos( ) =cos( )cos( ) +sin( )sin( )cos( )( )2 Corollary.

3 Given the Spherical triangleAB with opposing sides , , and , we have the following:sin( )cos(B) =cos( )sin( ) sin( )cos( )cos(A)( )Proof:Extend the side to 2radians as in Figure 3:Figure 3 Using The Spherical Law of Cosines, there are two ways of computingcos( ):cos( ) =cos( )cos( /2) +sin( )sin( /2)cos(B)( )=sin( )cos(B)( )cos( ) =cos( )cos( /2 c) +sin( )sin( /2 c)cos( A)( )=cos( )sin( ) sin( )cos( )cos(A)( )Equating ( ) and ( ), we get the corollary:sin( )cos(B) =cos( )sin( ) sin( )cos( )cos(A)( )32 Duality: Equators and PolesFor every great circle, there are two antipodal points whichare 2radians from every point on that greatcircle. Call these the poles of the great circle. Similarly,for each pair of antipodal points on a sphere, thereis a great circle, every point of which is 2radians from the pair.

4 Call this great circle the equator of theseantipodal points. The line containing the poles is perpendicular to the plane containing the equator. Thus,a central plane contains both poles if and only if it is perpendicular to the equatorial plane. Therefore,any great circle containing a pole is perpendicular to the equator, and any great circle perpendicular to theequator contains both 4: Semilunar TriangleBCis an Arc of the Equator for the PoleAIn Figure 4, BACis the angle between the plane containingABand the plane containingAC. As isevident in the view from aboveA, the length ofBCis the same as the size of (Semilune):A triangle in which one of the vertices is a pole of the opposing side is calleda semilunar triangle, or a described above, the angle at the pole has the same measureas the opposing side.

5 All of the other sidesand angles measure (Semilunar Lemma):If any two parts, a part being a side or an angle, of a sphericaltrianglemeasure 2radians, the triangle is a :There are four cases:1. two right sides2. two right angles3. opposing right side and right angle4. adjacent right side and right angleWe will handle these cases in 1 (two right sides):Suppose bothABandAChave a length of 2radians. The Spherical Law of Cosines sayscos(BC) =cos(AB)cos(AC) +sin(AB)sin(AC)cos( BAC)( )=cos( 2)cos( 2) +sin( 2)sin( 2)cos( BAC)( )=cos( BAC)( )Thus BACand opposing sideBCare equal. Furthermore,cos(AC) =cos(AB)cos(BC) +sin(AB)sin(BC)cos( ABC)( )cos( 2) =cos( 2)cos(BC) +sin( 2)sin(BC)cos( ABC)( )0 =sin(BC)cos( ABC)( )SinceBCis between 0 andpiradians,sin(BC)6= 0; thus,cos( ABC) = 0, and ABCmust be a similar argument, ACBmust also be 2 (two right angles):Suppose both ABCand ACBare right angles.

6 The Spherical Law of Cosines says thatcos(AC) =cos(AB)cos(BC) +sin(AB)sin(BC)cos( ABC)( )=cos(AB)cos(BC) +sin(AB)sin(BC)cos( 2)( )=cos(AB)cos(BC)( )Similarly,cos(AB) =cos(AC)cos(BC). Plugging this formula forcos(AB) into equation ( ), we getcos(AC) =cos(AC)cos2(BC)( )Subtracting the right side of equation ( ) from both sidesyieldscos(AC)sin2(BC) = 0( )SinceBCis between 0 and radians,sin(BC)6= 0. Therefore,cos(AC) = 0, andACis 2radians. By thesame argument,ABis also 2radians. Now apply Case 3 (opposing right side and right angle):Suppose both ABCandACmeasure 2radians. equation ( ) says thatcos(AC) =cos(AB)cos(BC)( )0 =cos(AB)cos(BC)( )Therefore, one ofABorBCmust be 2radians, and we are back to Case 4 (adjacent right side and right angle):Suppose both ABCandABmeasure 2radians. The Spherical Law of Cosines says thatcos(AC) =cos(AB)cos(BC) +sin(AB)sin(BC)cos( ABC)( )=cos( 2)cos(BC) +sin( 2)sin(BC)cos( 2)( )= 0( )Thus,ACis 2radians, and we are back to Case Dual TrianglesDefinition (Dual Triangle):Given a Spherical triangleABC, let4A B C be the triangle whosevertices are the poles of the sides opposite the corresponding vertices of4 ABCin the same hemisphere as4 ABC( is on the same side ofBCasA, etc.)

7 4A B C is the dual in Figure 5, leta,b, andcbe the sides oppositeA,B, andCrespectively, anda ,b , andc the sidesoppositeA ,B , andC . SinceA ,B , andC are the poles ofa,b, andc, all the red arcs measure 2radians. By construction,4 ABC ,4AB C, and4A BCare semilunes. However, by Lemma , so are4A B C,4A BC , and4AB C . Thus, the vertices of4 ABCare poles of the sides of4A B C , in theproper hemispheres. Therefore,4 ABCis the dual of4A B C .Figure 5: Dual TrianglesTheorem (Angle and Side Duality):The measure of an angle in a Spherical triangle and the lengthof the corresponding side in its dual are :Given4 ABC, let4A B C be its dual as constructed above. By the duality of the construction, weneed only consider one side and the angle at its corresponding pole, which is a vertex of the dual ACBandc in Figure 5.

8 As noted above,4A CB ,4AB C, and4A BCare semilunes. Thus, A CB andc are the same size. Furthermore, A CBand ACB are right angles. Therefore,c + ACB= A CB + ACB( )= A CB+ ACB ( )= 2+ 2( )= ( )Thus, we have that sidec and ACBare The Spherical Law of Cosines to the dual of a Spherical triangle, we get6 Theorem (The Law of Cosines for Angles):Given a Spherical triangle with two anglesAandBand the side between them, we can compute the cosine of opposite angle, , withcos( ) = cos(A)cos(B) +sin(A)sin(B)cos( )( )Proof:Consider4A B , the dual of4AB , with sides , , and . Apply The Spherical Law of Cosinesto compute :cos( ) =cos( )cos( ) +sin( )sin( )cos( )( )Use Theorem to replace each angle and side with the supplement of the corresponding side and angle inthe dualcos( ) =cos( A)cos( B) +sin( A)sin( B)cos( )( )Sincecos( x) = cos(x) andsin( x) =sin(x), this becomescos( ) = cos(A)cos(B) +sin(A)sin(B)cos( )( )Theorem (Incircle and Circumcircle Duality):The incenter of a Spherical triangle is the circum-center of its dual.

9 The inradius of a Spherical triangle is the complement of the the circumradius of :Given4 ABC, as in Figure 6, letGbe the center of its incircle, andD,E, andFbe the points oftangency of the incircle with sidesBC,AC, andABrespectively. Let4A B C be the dual 6: Incircle and Dual CircumcircleSince any radius of a circle is perpendicular to the circle,GDis perpendicular toBC. Therefore, if weextendDG, it passes throughA , the pole ofBC, andDA has length 2. The same is true for the otherpoints of tangency. Thus,GA ,GB , andGC are complementary tor, the inradius of4 ABC, and hence,equal. We can then conclude that the incenter of4 ABCis the circumcenter of4A B C and the inradiusof4 ABCis complementary to the circumradius of4A B C .74 Right Spherical TrianglesAs in plane Trigonometry , many facts about Spherical triangles can be derived using right Spherical (Projecting Right Angles):Projecting a right Spherical triangle onto a plane tangent toany of its vertices preserves the right :The case of projecting at the right angle is trivial.

10 Therefore, consider the right triangleABCinFigure 7 Construct the right triangleADC, congruent toABC, only reflected across sideAC. Made of two rightangles, BCDis a straight angle; thus,Cis onBD. LetB ,C , andD be the projections ofB,C, andDonto the plane tangent atA; thus,C is onB D . Since4 ABCis congruent to4 ADC,4AD C iscongruent to4AB C ; that is,AC =AC ,AB =tan(AB) =tan(AD) =AD , and B AC = BAC= DAC= D AC . Therefore, AC B = AC D , yet sinceC is onB D , AC B and AC D aresupplementary. Thus, each is a right says that right angles are preserved when projecting a Spherical triangle onto a plane tangentat any vertex of the given triangle. Corollary tells whathappens to other sizes of angles in a (Projecting Angles):Given a Spherical triangle projected onto a plane tangent atoneangle, the tangent of the projection of any other angle in thetriangle is the tangent of the correspondingspherical angle times the cosine of the edge connecting the :Consider4 ABDin Figure 7.