Spherical Trigonometry.docx

Spherical Triangle Any section made by a cutting plane that is circle. A great circle is formed when passes through the center of the sphere. triangle bounded by arc of great circles

passes through a sphere the cutting plane Spherical triangle is a of a sphere.

Note that for spherical triangles, sides a, b, and c are usually in angular units. And like plane triangles, angles A, B, and C are also in angular units. Sum of interior angles of spherical triangle The sum of the interior angles of a spherical triangle is greater than 180° and less than 540°. 180∘<(A+B+C)<540∘180∘<(A+B+C)<540∘

Area of spherical triangle The area of a spherical triangle on the surface of the sphere of radius R is given by the formula A=πR2E180∘A=πR2E180∘ Where E is the spherical excess in degrees. Spherical excess

E=A+B+C−180∘E=A+B+C−180∘ or

tan14E=tan12s tan12(s−a) tan12(s−b) tan12(s−c)−−−−−−−−−−−− −−−−−−−−−−−−−−−−−−−−−−−−√tan⁡14E=tan⁡12s tan⁡12(s−a) tan⁡12 (s−b) tan⁡12(s−c) Where s=12(a+b+c)s=12(a+b+c) Spherical defect

D=360∘−(a+b+c)D=360∘−(a+b+c)

Note: In spherical trigonometry, earth is assumed to be a perfect sphere. One minute (0° 1') of arc from the center of the earth has a distance equivalent to one (1) nautical mile (6080 feet) on the arc of great circle on the surface of the earth. 1 minute of arc = 1 nautical mile 1 nautical mile = 6080 feet 1 statute mile = 5280 feet 1 knot = 1 nautical mile per hour Right Spherical Triangle Solution of right spherical triangle With any two quantities given (three quantities if the right angle is counted), any right spherical triangle can be solved by following the Napier’s rules. The rules are aided with the Napier’s circle. In Napier’s circle, the sides and angle of the triangle are written in consecutive order (not including the right angle), and complimentary angles are taken for quantities opposite the right angle.

A¯=90∘−AA¯=90∘−A B¯=90∘−BB¯=90∘−B c¯=90∘−cc¯=90∘−c Napier’s Rules SIN-COOP Rule In the Napier’s circle, the sine of any middle part is equal to product of the cosines of its opposite parts. If we take aa as the middle part, its opposite parts are c¯c¯ and A¯A¯, then by sin-coop rule sina=cosc¯ cosA¯sin⁡a=cos⁡c¯ cos⁡A¯ sina=cos(90∘−c) cos(90∘−A)sin⁡a=cos⁡(90∘−c) cos⁡(90∘−A) sina=sinc sinAsin⁡a=sin⁡c sin⁡A SIN-TAAD Rule In the Napier’s circle, the sine of any middle part is equal to the product of the tangents of its adjacent parts. If we take aa as the middle part, the adjacent parts are bb and B¯B¯, then by sin-taad rule sinA=tanb tanB¯sin⁡A=tan⁡b tan⁡B¯ sinA=tanb tan(90∘−B)sin⁡A=tan⁡b tan⁡(90∘−B) Spherical triangle can have one or two or three 90° interior angle. Spherical triangle is said to be right if only one of its

included angle is equal to 90°. Triangles with more than one 90° angle are oblique.

Right Spherical Triangle Problem Solve for the spherical triangle whose parts are a = 73°, b = 62°, and C = 90°. Encircled the given parts for easy reference

To solve for angle A, use SIN-TAAD rule for b sinb=tanA¯ tanasin⁡b=tan⁡A¯ tan⁡a sin62∘=tan(90∘−A) tan73∘sin⁡62∘=tan⁡(90∘−A) tan⁡73∘ tan(90∘−A)=sin62∘tan73∘tan⁡(90∘−A)=sin⁡62∘tan⁡73∘ 90∘−A=15.11∘90∘−A=15.11∘ A=74.89∘A=74.89∘

answer

To solve for side c, use SIN-COOP rule for c¯c¯ sinc¯=cosa cosbsin⁡c¯=cos⁡a cos⁡b sin(90∘−c)=cos73∘ cos62∘sin⁡(90∘−c)=cos⁡73∘ cos⁡62∘ 90∘−c=7.89∘90∘−c=7.89∘ c=82.11∘c=82.11∘

answer

To solve for angle B, use SIN-TAAD rule for a sina=tanb tanB¯sin⁡a=tan⁡b tan⁡B¯ sin73∘=tan62∘ tan(90∘−B)sin⁡73∘=tan⁡62∘ tan⁡(90∘−B) tan(90∘−B)=sin73∘tan62∘tan(90∘−B)=sin⁡73∘tan⁡62∘ 90∘−B=26.95∘90∘−B=26.95∘ B=63.05∘B=63.05∘

answer