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Spherical TrigonometryIntroductory Theorems870DefinitionsPlanes through the centre of a sphere intersect the surface in great circles; other planes intersect it in small circles. Unless otherwise stated, all arcs are measured on great circles.The poles of a great circle are the extremities of the diameter perpendicular to its plane. The sides 𝑎, 𝑏, 𝑐 of a spherical triangle are the arcs of great circles 𝐵𝐶, 𝐶𝐴, 𝐴𝐵 on a sphere of radius unity; and the angles 𝐴, 𝐵, 𝐶 are the angles between the tangents to the sides at the vertices, or the angles between the planes of the great circles. The centre of the sphere will be denoted by 𝑂. The polar triangle of a spherical triangle 𝐴𝐵𝐶 has for its angular points 𝐴′, 𝐵′, 𝐶′, the poles of the sides 𝐵𝐶, 𝐶𝐴, 𝐴𝐵 of the primitive triangle in the directions of 𝐴, 𝐵, 𝐶 respectively (since each great circle has two poles). The sides of 𝐴′, 𝐵′, 𝐶′ are denoted by 𝑎′, 𝑏′, 𝑐′. 871 The sides and angles of the polar triangle are respectively the supplements of the angles and sides of the primitive triangle; that is, 𝑎′+𝐴=𝑏′+𝐵=𝑐′+𝐶=𝜋 𝑎+𝐴′=𝑏+𝐵′=𝑐+𝐶′=𝜋 Let 𝐵𝐶 produced cut the sides 𝐴′𝐵′, 𝐶′𝐴′ in 𝐺, 𝐻. 𝐵 is the pole of 𝐴′𝐶′, therefore 𝐵𝐻= 𝜋2. Similarly 𝐶𝐺= 𝜋2, therefore, by addition, 𝑎+𝐺𝐻=𝜋 and 𝐺𝐻=𝐴′, because 𝐴′ is the pole of 𝐵𝐶. The polar diagram of a spherical polygon is formed in the same way, and the same relations subsist between the sides and angles of the two figures. Rule: Hence, any equation between the sides and angles of a sphrical triangle porduces a supplementary equation by changing 𝑎 into 𝜋−𝐴 and 𝐴 into 𝜋−𝑎, ⋯ 872 The centre of the inscribed circle, radius 𝑟, is also the centre of the circumscribed circle, radius 𝑅′, of the polar triangle, and 𝑟+𝑅′= 12𝜋. Proof: In the last figure, let 𝑂 be the centre of the inscribed circle of 𝐴𝐵𝐶; then 𝑂𝐷, the perpendicular on 𝐵𝐶, passes through 𝐴′, the pole of 𝐵𝐶. Also, 𝑂𝐷=𝑟; therefore 𝑂𝐴′= 12𝜋−𝑟. similarly 𝑂𝐵′=𝑂𝐶′= 12𝜋−𝑟; therefore 𝑂 is the centre of the circumscribed circle of 𝐴′𝐵′𝐶′, and 𝑟+𝑅′= 12𝜋. 873 The sine of the arc joining a point on the circumference of a small circle with the pole of a parallel great circle, is equal to the ratio of the circumferences or corresponding arcs of the two circles. For it is equal to the radius of the small circle divided by the radius of the sphere; that is, by the radius of the great circle. 874 Two sides of a triangle are greater than the third. (By XI. 20) 875 The sides of a triangle are together less than the circumference of a great circle. (By XI. 21) 876 The angles of a triangel are together greater than two right angles. For 𝜋−𝐴+𝜋−𝐵+𝜋−𝐶 is <2𝜋, by (875) and the polar triangle. 877 If two sides of a triangle are equal, the opposite angles are equall. (By the geometrical proof in (894)) 878 If two angles of a triangle are equal, the opposite sides are equal. (By the polar triangle and (877)) 879 The greater angle of a triangle has the greater side opposite to it. Proof: If 𝐵 be >𝐴, draw the arc 𝐵𝐷 meeting 𝐴𝐶 in 𝐷, and make ∠𝐴𝐵𝐷=𝐴, therefore 𝐵𝐷=𝐴𝐷; but 𝐵𝐷+𝐷𝐶>𝐵𝐶, therefore 𝐴𝐶>𝐵𝐶. 880 The greater side of a triangle has the greater angle opposite to it. (By the polar triangle and (879)) RightAngled Triangles881 Napier's Rules: In the triangle 𝐴𝐵𝐶 let 𝐶 be a right angle, then 𝑎, (12𝜋−𝐵), ( 12𝜋−𝑐), ( 12𝜋−𝐴), and 𝑏, are called the five circular parts. Taking any part for middle part, Napier's rules are: I. sine of middle part = product of tangents of adjacent parts. II. sine of middle part = product of cosines of opposite parts. In applying the rules we can take 𝐴, 𝐵, 𝐶 instead of their complements, and change sine into Proof: From any point 𝑃 in 𝑂𝐴, draw 𝑃𝑅 perpendicular to 𝑂𝐶, and 𝑅𝑄 to 𝑂𝐵; therefore 𝑃𝑅𝑄 is a right angle; therefore 𝑂𝐵 is perpendicular to 𝑃𝑅 and 𝑄𝑅, and therefore to 𝑃𝑄. Then prove any formula by proportion from the triangles of the tetrahedron 𝑂𝑃𝑄𝑅, which are all rightangled. Otherwise, prove by the formula for obliqueangled triangles. ObliqueAngled Triangles882If 𝐴𝐵 and 𝐴𝐶 are both > 𝜋2, produce them to meet in 𝐴′, the pole of 𝐴, and employ the triangle 𝐴′𝐵𝐶. If 𝐴𝐵 alone be > 𝜋2, produce 𝐵𝐴 to meet 𝐵𝐶. 883 The supplementary formula, by (871), is 𝐴2=
𝐴2=
𝐴2=
12(𝑎+𝑏+𝑐) Proof: 𝐴2= 12(1− 𝐴2. 887 The supplementary formula are obtained in a similar way, or by the rule in (871). They are 𝑎2=
𝑎2= − 𝑎2= − 12(𝐴+𝐵+𝐶) 890 Let 𝜎= = 12 1+2891 Then the supplementary form, by (871), is ∑= −= 12 1−2892 2𝜎, 2∑By 𝐴2 𝐴2and (884, 885),⋯ 893 The following rules will produce the ten formula (884 to 892): I. Write 𝐴2, 𝐴2, 𝐴2, II. To obtain the supplementary forms of the five results, transpose large and small letters everywhere, and transpose = = Proof: By (882). Otherwise, in the figure of 882, draw 𝑃𝑁 perpendicular to 𝐵𝑂𝐶, and 𝑁𝑅, 𝑁𝑆 to 𝑂𝐵, 𝑂𝐶. Prove 𝑃𝑅𝑂 and 𝑃𝑆𝑂 right angles by I. 47 and therefore 𝑃𝑁=𝑂𝑃 Rule: Product of cosines of middle parts = Proof: In the formula for . Napier's Formula89612(𝐴−𝐵)=
𝐶21 12(𝐴+𝐵)=
𝐶22 12(𝑎−𝑏)=
𝐶23 12(𝑎+𝑏)=
𝐶24 Rule: In the value of 12(𝐴−𝐵) change 12(𝐴+𝐵). To obtain [3] and [4] from [1] and [2], transpose sides and angles, and change Proof: In the values of , and transform by (670672) Gauss's Formula897= 1 = 2 = 3 = 4 From any of these formula the others may be obtained by the following rule: Rule: Change the sign of the letter 𝐵 (large or small) on one side of the equation, and write Proof: Take Spherical Triangle and Circle898 Let 𝑟 be the radius of the inscribed circle of 𝐴𝐵𝐶; 𝑟_{𝑎} the radius of the escribed circle touching the side 𝑎, and 𝑅, 𝑅_{𝑎} the radii of the circumscribed circles; then
Proof: The first value is found from the rightangled triangle 𝑂𝐴𝐹, in which 𝐴𝐹=𝑠−𝑎. The other values by (884892).
899
Proof: From the rightangled triangle 𝑂′𝐴𝐹, in which 𝐴𝐹′=𝑠. Note: The first two values of
Proof: The first value from the rightangled triangle 𝑂𝐵𝐷, in which ∠𝑂𝐵𝐷=𝑆−𝐴. The other values by the formula (887802)
901
Proof: From the rightangled triangle 𝑂′𝐵𝐷, in which ∠𝑂𝐵𝐷=𝜋−𝑆.
Spherical Areas902 area of 𝐴𝐵𝐶=(𝐴+𝐵+𝐶−𝜋)𝑟^{2}=𝐸𝑟^{2}, where 𝐸=𝐴+𝐵+𝐶−𝜋, the spherical excess Proof: By adding the three lunes 𝐴𝐵𝐷𝐶, 𝐵𝐶𝐸𝐴, 𝐶𝐴𝐹𝐵, and observing that 𝐴𝐵𝐹=𝐶𝐷𝐸, we get2𝜋𝑟^{2}=2𝜋𝑟^{2}+2𝐴𝐵𝐶 Area of Spherical Polygon903 𝑛 being the number of sides, Area={Interior Angles−(𝑛−2)𝜋}𝑟^{2} ={2𝜋−Exterior Angles}𝑟^{2} ={2𝜋−sides of Polar Diagram}𝑟^{2} The last value holds for curvilinear area in the limit.Proof: By joining the vertices with an interior point, and adding the areas of the spherical triangles so formed. Cagnoli's Theorem90412𝐸= Proof: Expand 12(𝐴+𝐵)− 12(𝜋−𝐶)] by (628), and transform by Gauss's equations (897i, iii) and (669, 890). Llhuillier's Theorem90514𝐸= 14(𝐴+𝐵−𝐶+𝜋) and reduce by (667, 668), then eliminate 12(𝐴+𝐵) by Gauss's formula (897 i, iii). Transform by (672, 673), and substitute from (886). PolyhedronsLet the number of faces, solid angles, and edges, of any polyhedron be 𝐹, 𝑆, 𝐸; then 906 𝐹+𝑆=𝐸+2 Proof: Project the polyhedron upon an internal sphere. Let 𝑚=number of sides, and 𝑠=sum of angles of one of the spherical polygons so formed. Then its area={𝑠−(𝑚−2)𝜋}𝑟^{2}, by (903). Sum this for all the polygons, and equate to 4𝜋𝑟^{2}.The Five Regular solidsLet 𝑚 be the number of sides in each face, 𝑛 the number of plane angles in each solid angle; therefore 907 𝑚𝐹=𝑛𝑆=2𝐸 From these equations and (906), find 𝐹, 𝑆, and 𝐸 in terms of 𝑚 and 𝑛, thus,1𝐹= 𝑚2 , 1𝑆= 𝑛2 , 1𝐸= In order that 𝐹, 𝑆, and 𝐸 may be positive, we must
Regular Solids𝑚𝑛𝐹𝑆𝐸
Tetrahedron33446
Hexahedron436812
Octahedron348612
Dodecahedron35201230
Icosahedron35201230
908
The sum of all the plane angles of any polyhedron
=2𝜋(𝑆−2)
Or, Four right angles for every vertex less eight right angles.
909
If 𝐼 be the angle between two adjacent faces of a regular polyhedron.
12𝐼= 𝜋𝑛÷ 𝜋𝑚Proof: Let 𝑃𝑄=𝑎 be the edge, and 𝑆 the centre of a face, 𝑇 the middle point of 𝑃𝑄, 𝑂 the centre of the inscribed and circumscribed spheres, 𝐴𝐵𝐶 the projection of 𝑃𝑆𝑇 upon a concentric sphere. In this spherical triangle. 𝐶= 𝜋2, 𝐴= 𝜋𝑛, 𝐴= 𝜋𝑚=𝑃𝑆𝑇, Also 𝑆𝑇𝑂= 12𝐼 Now, by (881, ii), 𝜋𝑛= 𝜋𝑚 12𝐼Q.e.d. 910 If 𝑟, 𝑅 be the radii of the inscribed and circumscribed spheres of a regular polyhedron, 𝑟= 𝑎2 12𝐼 𝜋𝑚, 𝑅= 𝑎2 12𝐼 𝜋𝑛Proof: In the above figure, 𝑂𝑆=𝑟, 𝑂𝑃=𝑅, 𝑃𝑇= 𝑎2; and 𝑂𝑆=𝑃𝑇 𝜋𝑚 12𝐼. Also 𝑂𝑃=𝑃𝑇 12𝐼 𝜋𝑛; therefore, ⋯ Sources and Referenceshttps://archive.org/details/synopsisofelementaryresultsinpureandappliedmathematicspdfdrive©sideway ID: 210900014 Last Updated: 9/14/2021 Revision: 0 Ref: References
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