Spherical Trigonometry

ENGTRIG: LECTURE # 4.2 Spherical Trigonometry
Spherical Trigonometry

Engr. Christian Pangilinan

Areas of a Spherical Triangle
A=

π R2 E
180o E R

E = A + B + C − 180o

Where:

spherical excess radius of the sphere

Spherical Triangles Part of the surface of the sphere bounded by three arcs of three great circles Right Spherical Triangle – a spherical triangle containing at least one right angle

If the sides are known instead of the angles, then L’Huiller’s Formula can be used to solve for the spherical excess

1 s  s − a  s − b  s−c   tan E =  tan   tan   tan  tan  2 2  2  2  2   a+b+c Where: s= 2
Solutions to Right Spherical Triangles (C = 90o)

Sides a, b, c are based on the corresponding arc lengths S = rθ that is based on its corresponding interior angles S a = rθ BOC ; Sb = rθ AOC ; Sc = rθ AOB and that a + b + c < 360o Angles A, B, C of a spherical triangle are measured by the corresponding dihedral angles of the trihedral angle A : B − OA − C ; B : A − OB − C and
C : A − OC − B

Napier’s Rules: 1. The sine of any middle part is equal to the product of the tangents of the adjacent parts 2. The sine of any middle part is equal to the product of the opposite parts *“co” indicates complement

s i n ( middle ) = product of t a n ( adjacent ) s i n ( middle ) = product of c o s ( opposite )

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Angles are measured between the tangents at the point (A) to the arcs of the other points (B and C) and that 180o < A + B + C < 540o

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ENGTRIG: LECTURE # 4.2 Spherical Trigonometry
Solve for the remaining angles and sides and the area given the ff. conditions: 1. C = 90o ; 2. C = 90o ; 3. C = 90o ;
A = 70o ; c = 75o ; R = 5 A = 120o ; a = 100o ; R = 7 A = 42o ; b = 30o ; R = 3

Engr. Christian Pangilinan

Solve for the remaining angles and sides and the area given the ff. conditions: 1. a = 125o ; b = 53o ; c = 98o 2. A = 43o ; B = 136o ; C = 65o 3. a = 56o ; b = 20o ; C = 114o 4. A = 77 o ;