Spherical Trigonometry

LessonAbout 1.5 h2 of 2 in Plane and Spherical Trigonometry

Spherical Trigonometry

Fundamentals of Spherical Triangles

A spherical triangle is a figure on the surface of a sphere bounded by three arcs of great circles. Unlike plane triangles, the sum of the interior angles of a spherical triangle always exceeds \(180^\circ\) and is less than or equal to \(540^\circ\).

Angle sum range for a spherical triangle:
\[ 180^\circ < A + B + C \le 540^\circ \]
  • Bounded by three great circle arcs on sphere
  • Sum of interior angles exceeds 180 degrees
  • Maximum sum of interior angles is 540 degrees

Right Spherical Triangles and Napier’s Rules

A right spherical triangle contains one right angle, usually designated as angle \(C = 90^\circ\). The remaining five elements—two legs \(a\) and \(b\), hypotenuse \(c\), and interior angles \(A\) and \(B\)—are solved using Napier’s Rules on a five-part circle.

Napier’s circle arranges the five parts in circular order, omitting the right angle \(C\): side \(a\), side \(b\), co-A (\(90^\circ - A\)), co-c (\(90^\circ - c\)), and co-B (\(90^\circ - B\)). Any part selected as the middle part relates to its adjacent parts and opposite parts.

Tan-Ad Rule (SIN-TAAD): Sine of middle part equals product of tangents of adjacent parts:
\[ \sin(\text{middle}) = \tan(\text{adjacent}_1) \tan(\text{adjacent}_2) \]
Co-Op Rule (SIN-COOP): Sine of middle part equals product of cosines of opposite parts:
\[ \sin(\text{middle}) = \cos(\text{opposite}_1) \cos(\text{opposite}_2) \]
Fundamental right triangle relations:
\[ \sin a = \tan(\text{co-}B) \tan b = \cot B \tan b \]
\[ \sin a = \cos(\text{co-}c) \cos(\text{co-}A) = \sin c \sin A \]
\[ \sin b = \cos(\text{co-}c) \cos(\text{co-}B) = \sin c \sin B \]
\[ \cos c = \cos a \cos b \]
\[ \cos A = \tan b \cot c \]

Worked Example 1: Finding missing side given leg and hypotenuse

In right spherical triangle \(ABC\) with \(C = 90^\circ\), \(a = 50^\circ\) and \(c = 80^\circ\). Solve for side \(b\).

Solution: Applying the Co-Op rule with \(\text{co-}c\) as middle part gives \(\sin(\text{co-}c) = \cos a \cos b\), or \(\cos c = \cos a \cos b\).

\[ \cos b = \dfrac{\cos c}{\cos a} = \dfrac{\cos 80^\circ}{\cos 50^\circ} = \dfrac{0.17365}{0.64279} = 0.27015 \]
\[ b = \arccos(0.27015) = 74.33^\circ \]

Worked Example 2: Finding hypotenuse given two legs

In right spherical triangle \(ABC\) with \(C = 90^\circ\), \(a = 82^\circ\) and \(b = 62^\circ\). Find hypotenuse \(c\).

Solution: Using the cosine hypotenuse formula \(\cos c = \cos a \cos b\):

\[ \cos c = \cos 82^\circ \cos 62^\circ = (0.13917)(0.46947) = 0.06534 \]
\[ c = \arccos(0.06534) = 86.25^\circ = 86^\circ 15' \]

Worked Example 3: Isosceles spherical triangle solved by perpendicular bisector

In isosceles spherical triangle \(ABC\), \(A = B = 80^\circ\) and \(a = b = 89^\circ\). Solve for side \(c\).

Solution: Drop a perpendicular altitude \(h\) from vertex \(C\) to bisect side \(c\) into two equal parts \(c/2\). In the right spherical triangle formed with hypotenuse \(b = 89^\circ\) and angle \(A = 80^\circ\):

\[ \sin(\text{co-}A) = \tan(c/2) \tan(\text{co-}b) \implies \cos A = \tan(c/2) \cot b \]
\[ \tan(c/2) = \cos A \tan b = \cos 80^\circ \tan 89^\circ = (0.17365)(57.2900) = 9.948 \]
\[ c/2 = \arctan(9.948) = 84.26^\circ \implies c = 2(84.26^\circ) = 168.52^\circ = 168^\circ 31' \]

Worked Example 4: Finding angle of isosceles spherical triangle given sides

In isosceles spherical triangle \(ABC\), \(b = c = 54^\circ 28'\) and \(a = 92^\circ 30'\). Find angle \(B\).

Solution: Bisect side \(a\) into \(a/2 = 46^\circ 15'\). In the resulting right spherical triangle with leg \(a/2\), hypotenuse \(b\), and angle \(B\):

\[ \sin(\text{co-}B) = \tan(a/2) \tan(\text{co-}b) \implies \cos B = \tan(a/2) \cot b \]
\[ \cos B = \dfrac{\tan 46^\circ 15'}{\tan 54^\circ 28'} = \dfrac{1.0445}{1.3998} = 0.7462 \implies B = 41^\circ 45' \]

Oblique Spherical Triangles

An oblique spherical triangle contains no right angle. Its solution requires the Law of Sines, Law of Cosines for Sides, Law of Cosines for Angles, or Napier’s Analogies.

Law of Sines:
\[ \dfrac{\sin a}{\sin A} = \dfrac{\sin b}{\sin B} = \dfrac{\sin c}{\sin C} \]
Law of Cosines for Sides:
\[ \cos a = \cos b \cos c + \sin b \sin c \cos A \]
\[ \cos b = \cos a \cos c + \sin a \sin c \cos B \]
\[ \cos c = \cos a \cos b + \sin a \sin b \cos C \]
Law of Cosines for Angles:
\[ \cos A = -\cos B \cos C + \sin B \sin C \cos a \]
\[ \cos B = -\cos A \cos C + \sin A \sin C \cos b \]
\[ \cos C = -\cos A \cos B + \sin A \sin B \cos c \]
Napier’s Analogies:
\[ \dfrac{\sin\frac{1}{2}(A-B)}{\sin\frac{1}{2}(A+B)} = \dfrac{\tan\frac{1}{2}(a-b)}{\tan\frac{1}{2}c} \]
\[ \dfrac{\sin\frac{1}{2}(a-b)}{\sin\frac{1}{2}(a+b)} = \dfrac{\tan\frac{1}{2}(A-B)}{\cot\frac{1}{2}C} \]

Worked Example 5: Two sides and included angle (SAS)

In spherical triangle \(ABC\), \(a = 106^\circ 25'\), \(c = 42^\circ 16'\), and \(B = 114^\circ 53'\). Find angle \(A\).

Solution: First calculate side \(b\) using Law of Cosines for sides:

\[ \cos b = \cos 106^\circ 25' \cos 42^\circ 16' + \sin 106^\circ 25' \sin 42^\circ 16' \cos 114^\circ 53' \]
\[ \cos b = (-0.2826)(0.7400) + (0.9592)(0.6726)(-0.4208) = -0.2091 - 0.2715 = -0.4806 \implies b = 118^\circ 43' \]

Next find angle \(A\) using Law of Sines:

\[ \sin A = \dfrac{\sin a \sin B}{\sin b} = \dfrac{\sin 106^\circ 25' \sin 114^\circ 53'}{\sin 118^\circ 43'} = \dfrac{(0.9592)(0.9071)}{0.8770} = 0.9922 \]

Since side \(a > c\), angle \(A\) must be obtuse: \(A = 180^\circ - 82^\circ 51' = 97^\circ 09'\).

Worked Example 6: Two sides and one opposite angle (SSA)

In spherical triangle \(ABC\), \(a = 80^\circ\), \(c = 115^\circ\), and \(A = 72^\circ\). Solve for angle \(C\).

Solution: Using Law of Sines:

\[ \dfrac{\sin C}{\sin c} = \dfrac{\sin A}{\sin a} \implies \sin C = \dfrac{\sin 115^\circ \sin 72^\circ}{\sin 80^\circ} = \dfrac{(0.9063)(0.9511)}{0.9848} = 0.8752 \]

Since \(c > a\), angle \(C > A\), making \(C\) obtuse: \(C = 180^\circ - 61^\circ 04' = 118^\circ 56'\).

Worked Example 7: Three angles given (AAA)

In spherical triangle \(ABC\), \(A = 116^\circ 19'\), \(B = 55^\circ 30'\), and \(C = 80^\circ 37'\). Solve for side \(a\).

Solution: Apply Law of Cosines for angles:

\[ \cos A = -\cos B \cos C + \sin B \sin C \cos a \]
\[ \cos 116^\circ 19' = -\cos 55^\circ 30' \cos 80^\circ 37' + \sin 55^\circ 30' \sin 80^\circ 37' \cos a \]
\[ -0.4433 = -(0.5664)(0.1630) + (0.8241)(0.9866) \cos a \implies -0.4433 = -0.0923 + 0.8131 \cos a \]
\[ \cos a = \dfrac{-0.3510}{0.8131} = -0.4317 \implies a = 115.57^\circ \]

Spherical Excess and Area of Spherical Triangles

Spherical excess \(E\) is the amount by which the sum of the interior angles of a spherical triangle exceeds \(180^\circ\). The surface area of a spherical triangle is directly proportional to its spherical excess.

Spherical Excess formula:
\[ E = A + B + C - 180^\circ \]
L’Huilier’s Formula (given three sides):
\[ \tan\dfrac{E}{4} = \sqrt{\tan\dfrac{s}{2} \tan\dfrac{s-a}{2} \tan\dfrac{s-b}{2} \tan\dfrac{s-c}{2}} \]
where \(s = \dfrac{a + b + c}{2}\). Area of Spherical Triangle:
\[ A_{area} = \dfrac{\pi R^2 E}{180^\circ} \]

Worked Example 8: Area from three interior angles and sphere radius

Find the area of spherical triangle \(ABC\) with angles \(A = 93^\circ 40'\), \(B = 64^\circ 12'\), and \(C = 116^\circ 51'\) on a sphere of radius \(R = 100\text{ m}\).

Solution: Compute spherical excess \(E\):

\[ E = 93^\circ 40' + 64^\circ 12' + 116^\circ 51' - 180^\circ = 274^\circ 43' - 180^\circ = 94^\circ 43' = 94.7167^\circ \]

Compute surface area \(A_{area}\):

\[ A_{area} = \dfrac{\pi (100)^2 (94.7167^\circ)}{180^\circ} = 16,531\text{ m}^2 \]

Worked Example 9: Finding sphere radius from area and excess

A spherical triangle has an area of \(327.25\text{ km}^2\) and spherical excess \(E = 30^\circ\). Find the radius \(R\) of the sphere.

Solution: Substitute into area formula \(A_{area} = \dfrac{\pi R^2 E}{180^\circ}\):

\[ 327.25 = \dfrac{\pi R^2 (30^\circ)}{180^\circ} = \dfrac{\pi R^2}{6} \]
\[ R^2 = \dfrac{327.25 \times 6}{\pi} = 624.88 \implies R = \sqrt{624.88} = 25\text{ km} \]

Worked Example 10: Spherical excess from three sides

Determine the spherical excess \(E\) of triangle \(ABC\) given sides \(a = 56^\circ\), \(b = 65^\circ\), and \(c = 78^\circ\).

Solution: Find \(A\) using Law of Cosines for sides: \(\cos 56^\circ = \cos 65^\circ \cos 78^\circ + \sin 65^\circ \sin 78^\circ \cos A \implies A = 57^\circ 53'\).

Find \(B\) by Law of Sines: \(\sin B = \dfrac{\sin 65^\circ \sin 57^\circ 53'}{\sin 56^\circ} \implies B = 67^\circ 48'\).

The Law of Sines gives two possible angles for \(C\) (\(87^\circ 50'\) or \(92^\circ 10'\)), so find \(C\) with the Law of Cosines for angles instead. Since \(c\) is the largest side, \(C\) is the largest angle:

\[ \cos C = \dfrac{\cos c - \cos a \cos b}{\sin a \sin b} = \dfrac{\cos 78^\circ - \cos 56^\circ \cos 65^\circ}{\sin 56^\circ \sin 65^\circ} = \dfrac{0.2079 - 0.2363}{0.7513} = -0.0378 \implies C = 92^\circ 10' \]
\[ E = 57^\circ 53' + 67^\circ 48' + 92^\circ 10' - 180^\circ = 37^\circ 51' \]

Terrestrial Sphere and Navigation Applications

The Earth is modeled as a terrestrial sphere of radius \(R = 3959\text{ miles}\). Navigation calculations use great circles (meridians and equator) and parallels of latitude.

Terrestrial Constants and Conversion Factors:
\[ 1\text{ minute of arc } (1') \text{ on great circle} = 1\text{ Nautical Mile (NM)} \]
\[ 1\text{ NM} = 6080\text{ feet} \]
\[ 1\text{ statute mile} = 5280\text{ feet} \]
\[ 1^\circ \text{ on equator} = 60\text{ NM} = 4\text{ minutes of time} \]
\[ 15^\circ \text{ longitude difference} = 1\text{ hour time difference} \]
\[ 1\text{ knot} = 1\text{ NM per hour} \]
  • Meridians are great circles passing through poles
  • Parallels are circles parallel to equator
  • Latitude measured north or south from equator
  • Longitude measured east or west from prime meridian

Worked Example 11: Time difference from longitude

If Greenwich Mean Time (GMT) is 6:00 AM, determine the time at a location situated at \(30^\circ\text{ East}\) longitude.

Solution: Calculate time difference:

\[ \Delta t = \dfrac{30^\circ - 0^\circ}{15^\circ/\text{hr}} = 2\text{ hours} \]

Since the location is East of Greenwich, time is ahead: \(6:00\text{ AM} + 2\text{ hours} = 8:00\text{ AM}\).

Worked Example 12: Time difference between two cities

Find the time difference between Tokyo (\(139^\circ\text{ E}\)) and Manila (\(121^\circ\text{ E}\)).

Solution: Difference in longitude \(\Delta \lambda = 139^\circ - 121^\circ = 18^\circ\).

\[ \Delta t = \dfrac{18^\circ}{15^\circ/\text{hr}} = 1.2\text{ hours} = 1\text{ hour and } 12\text{ minutes} \]

Worked Example 13: Great circle distance and speed

A ship sails along a great circle path between \(A(140^\circ\text{ E}, 20^\circ\text{ N})\) and \(B(170^\circ\text{ E}, 10^\circ\text{ S})\) in 150 hours. Calculate distance in nautical miles and speed in knots.

Solution: The triangle formed by the North Pole and points \(A\) and \(B\) is oblique. Its two sides from the pole are the colatitudes \(90^\circ - 20^\circ = 70^\circ\) and \(90^\circ + 10^\circ = 100^\circ\), and the included angle at the pole is the longitude difference \(\Delta \lambda = 170^\circ - 140^\circ = 30^\circ\). Apply the Law of Cosines for sides:

\[ \cos c = \cos 70^\circ \cos 100^\circ + \sin 70^\circ \sin 100^\circ \cos 30^\circ \]
\[ \cos c = (0.34202)(-0.17365) + (0.93969)(0.98481)(0.86603) = -0.05939 + 0.80145 = 0.74206 \implies c = 42.09^\circ \]
\[ \text{Distance } AB = 42.09^\circ \times 60\text{ NM/deg} = 2525.7\text{ NM} \]
\[ \text{Speed } V = \dfrac{2525.7\text{ NM}}{150\text{ hrs}} = 16.84\text{ knots} \]

Worked Example 14: Great circle distance between global coordinates

Find the great circle distance between San Francisco (\(37^\circ 48'\text{ N}, 122^\circ 24'\text{ W}\)) and Manila (\(14^\circ 36'\text{ N}, 121^\circ 05'\text{ E}\)).

Solution: Difference in longitude \(\Delta \lambda = 360^\circ - (122^\circ 24' + 121^\circ 05') = 116^\circ 31'\). Co-latitudes are \(a = 90^\circ - 37^\circ 48' = 52^\circ 12'\) and \(b = 90^\circ - 14^\circ 36' = 75^\circ 24'\). By Law of Cosines for sides:

\[ \cos s = \cos 52^\circ 12' \cos 75^\circ 24' + \sin 52^\circ 12' \sin 75^\circ 24' \cos 116^\circ 31' \]
\[ \cos s = (0.6129)(0.2521) + (0.7902)(0.9677)(-0.4465) = 0.1545 - 0.3414 = -0.1869 \implies s = 100.77^\circ \]
\[ \text{Distance} = 100.77^\circ \times 60\text{ NM/deg} = 6046.2\text{ NM} \]

Formula Summary: Spherical Trigonometry

Formula Name Formula
Spherical Angle Sum Range \(180^\circ < A + B + C \le 540^\circ\)
Napier’s Tan-Ad Rule \(\sin(\text{middle}) = \tan(\text{adjacent}_1) \tan(\text{adjacent}_2)\)
Napier’s Co-Op Rule \(\sin(\text{middle}) = \cos(\text{opposite}_1) \cos(\text{opposite}_2)\)
Right Triangle Cosine of Hypotenuse \(\cos c = \cos a \cos b\)
Right Triangle Sine Law \(\sin a = \sin c \sin A\)
Right Triangle Cosine-Tangent Relation \(\cos A = \tan b \cot c\)
Spherical Law of Sines \(\dfrac{\sin a}{\sin A} = \dfrac{\sin b}{\sin B} = \dfrac{\sin c}{\sin C}\)
Spherical Law of Cosines for Sides \(\cos a = \cos b \cos c + \sin b \sin c \cos A\)
Spherical Law of Cosines for Angles \(\cos A = -\cos B \cos C + \sin B \sin C \cos a\)
Napier’s Analogies \(\dfrac{\sin\dfrac{1}{2}(A-B)}{\sin\dfrac{1}{2}(A+B)} = \dfrac{\tan\dfrac{1}{2}(a-b)}{\tan\dfrac{1}{2}c}\)
Spherical Excess \(E = A + B + C - 180^\circ\)
L’Huilier’s Formula \(\tan\dfrac{E}{4} = \sqrt{\tan\dfrac{s}{2} \tan\dfrac{s-a}{2} \tan\dfrac{s-b}{2} \tan\dfrac{s-c}{2}}\)
Spherical Area \(A_{area} = \dfrac{\pi R^2 E}{180^\circ}\)
Great Circle Distance \(d = c \times 60\)
Longitude Time Difference \(\Delta t = \dfrac{\Delta \lambda}{15^\circ/\text{hr}}\)