Spherical Trigonometry
DEFINITION OF TERMS
DEFINITION OF TERMS
GREAT CIRCLE
 a circle on the surface of a sphere, whose plane passes through the
center of the sphere
small circle
SPHERICAL ANGLE
 an angle formed by the intersection
of two great circles
2
 always bisects the sphere
3
1
great circle
great circle
DEFINITION OF TERMS
PROPERTIES OF SPHERICAL TRIANGLE
SPHERICAL TRIANGLE
 a triangle on the surface of the
sphere, formed by the intersection
of three Great circles
 The greater side has the greater angle
opposite to it
 the bounding arcs are called the
sides and the angles formed by
intersecting arcs are the angles
 The sum of the three sides of a
spherical triangle is less than 3600.
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 The sum of any two sides is greater
than the third
 The sum of the interior angles of a
spherical triangle is greater than 180°
but less than 5400.
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Spherical Trigonometry
RIGHT SPHERICAL TRIANGLES: is one with a right angle.

A
OBLIQUE SPHERICAL TRIANGLE
co = complement of
c
co-A
b
B
a
C
b
co-c
co-B
a
Note:
 none of its angles is 90°, or
sin co-A = cos A
cos co-A = sin A
tan co-A = cot A
 none of its two or three of its
angles equal 90°
NAPIER’S RULES:
Rule 1: (Sin-Ta-Ad Rule)
The sine of any middle part is equal to the product of the tangents of its adjacent parts.
Rule 2: (Sin-Co-Op Rule)
The sine of any middle part is equal to the product of the cosines of its opposite parts.
OBLIQUE SPHERICAL TRIANGLES
Sine Law:
sin a sin b sin c
=
=
sin A sin B sin C
1. Solve the unknown angles and side of the spherical
triangle whose given parts are:
a = 720 27’
b = 610 49’
C = 900
A
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b
co-c
Cosine Law 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
Cosine Law 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
co-A
Rule 2: (Sin-Co-Op Rule)
The sine of any middle part is equal to the
product of the cosines of its opposite parts.
c
B
co-B
a
b
a
C
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Spherical Trigonometry
1. Solve the unknown angles and side of the spherical
triangle whose given parts are:
a = 720 27’
b = 610 49’
C = 900
sin (co-c) = cos a cos b
cos c = cos a cos b
cos c = cos (72027’) cos (61049’)
A
c
B
b
co-c
co-B
a
c
B
sin (co-c) = cos a cos b
cos c = cos a cos b
cos c = cos (72027’) cos (61049’)
A
c
b
C
co-A
b
co-c
co-B
sin b = tan a cot A
tan 72027’
sin 61049’ =
tan A
A = 74.420
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c
B
b
a
C
a
c = 81.810
sin (co-c) = cos a cos b
cos c = cos a cos b
cos c = cos (72027’) cos (61049’)
A
a
b
Rule 1: (Sin-Ta-Ad Rule)
The sine of any middle part is equal to the
product of the tangents of its adjacent parts.
C
a
co-B
1. Solve the unknown angles and side of the spherical
triangle whose given parts are:
a = 720 27’
b = 610 49’
C = 900
c = 81.810
sin b = tan a tan (co-A)
b
co-A
co-c
sin (co-c) = cos a cos b
cos c = cos a cos b
cos c = cos (72027’) cos (61049’)
A
C
a
a
b
c = 81.810
1. Solve the unknown angles and side of the spherical
triangle whose given parts are:
a = 720 27’
b = 610 49’
C = 900
B
co-A
1. Solve the unknown angles and side of the spherical
triangle whose given parts are:
a = 720 27’
b = 610 49’
C = 900
co-A
b
co-c
co-B
a
c = 81.810
sin b = tan a tan (co-A)
sin a = tan b tan (co-B)
sin b = tan a cot A
tan 72027’
sin 61049’ =
tan A
sin a = tan b cot B
tan 61049’
sin 72027’ =
tan B
A = 74.420
B = 62.940
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Spherical Trigonometry
2. Solve the triangle whose given parts are:
A = B = 640 37’,
and b = 810 14’
C
A = B; Isosceles Spherical Triangle
C C
2 2
h
b
A c/2
c
c/2 B
co-A
c/2
sin (co-A) = tan (c/2) tan (co-b)
A c/2
cos A = tan (c/2) cot b
h
b
a = b = 81014’
C = 1400 48’
C
A. 18.280
C. 161.720
B. 20.190
D. 159.810
=
sin 152019’
sin C
sin 110033’
co-A
h
c/2
cos A = tan (c/2) cot b
sin (co-b) = tan (co-A) tan (co-C/2)
tan c/2
cos 64037’=
tan 81014’
cos b = cot A cot (C/2)
1
cos 81014’=
tan 64037’ tan (C/2)
c = 140.430
C = 144.380
Prime Meridian (Greenwich Line)
=
a
b
sin c
B
A
sin 140048’
*The greater side has the greater angle opposite to it.
c=
E = 600 + 800 + 1040 - 1800 = 640
A=
Latitude
161.720
4. Find the area of the spherical triangle whose angles are A = 600, B = 800, and C = 1040.
The radius of the sphere is 20cm.
E = spherical excess
= sum of angles - 180°
= A + B + C - 180°
Longitude
c
c = 1800 - 18.280
c = 18.280
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co-b
a = 1520 19’
and
sin c
co-C/2
sin (co-A) = tan (c/2) tan (co-b)
A c/2
A = 1100 33’,
πR E
A=
180
C
2
h
3. Given the parts of spherical triangle, solve for side C:
sin A
c
a
c/2 B
c = 140.430
A c/2
sin a
C C
2 2
h
b
tan c/2
cos 64037’=
tan 81014’
C
2
b
h
co-b
a = b = 81014’
a
2. Solve the triangle whose given parts are:
A = B = 640 37’,
and b = 810 14’
C
A = B; Isosceles Spherical Triangle
co-C/2
π 20 (64 )
180
A = 𝟒𝟒𝟔. 𝟖𝟎 𝐜𝐦𝟐
Equator
For long distances and large areas, the earth is considered a sphere,
called TERRESTRIAL SPHERE.
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Spherical Trigonometry
5. A plane left Manila (14°36’N, 121°5’E) and flew in the direction of S 32° E.
At what longitude will it cross the equator?
prime meridian
14°36’ = b
North Pole
●
C
●
●B
a
co-B
co-c
a
co-A
b
SINTAAD Rule
sin b = tan a tan co-A
sin 14°36’= tan a cot 32o
a = 8°57’4.11’’
Longitude of B:
= 121°05’ + 8°57’4.11’’
c
b
A = 32°
Equator
Longitude of B = 130°02’4.11’’
South Pole
7. Find the time it would take an airplane flying at a speed of 1200kph to fly along a
great circle route from Manila (14038’N, 12105’E) to Moscow (55045’N, 37037’E). Find
its course if it flew from Manila and from Moscow.
a = 900 - 55045’ = 34015’
b = 900 - 14038’ = 75022’
C = 121005’ - 37037’ = 83028’
A(14°38’N,121°5’E)
prime meridian
B(55°45’N,37°37’E)
a
North Pole
C
cos c = cos 34015’ cos 75022’ +
sin 34015’ sin 75022’ cos 83028’
B●
●
prime meridian
c = 74.290
●
A
●
Equator
V=
South Pole
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d
t
8262.50 km
1200 kph =
t
t = 6.89 hrs
cos c = cos a cos b + sin a sin b cos C
a
A
●
c
●
B(33°52’S,151°12’E)
cos c = cos 123052’ cos 75025’ +
sin 1230 52’ sin 75025’ cos 30013’
c = 56.350
●
●B
Equator
Note:
1 minute of arc = 1 nautical mile
1 nautical mile
60’
x
1’
10
Distance = 56.350 x
South Pole
Distance = 3381 nautical miles
7. Find the time it would take an airplane flying at a speed of 1200kph to fly along a
great circle route from Manila (14038’N, 12105’E) to Moscow (55045’N, 37037’E). Find
its course if it flew from Manila and from Moscow.
a = 900 – 550 45’ = 340 15’
A(14°38’N,121°5’E)
prime meridian
B(55°45’N,37°37’E)
a
North Pole
C
b
b = 900 – 140 38’ = 750 22’
C = 121005’ – 370 37’ = 830 28’
c = 74.290
B●
c
60’
10
1m
North Pole
C
A(14°35’N
,120°59’E)
NOTE:
1 minute of arc = 1 nautical mile
1 nautical mile = 6080 ft
6080 ft
x 1 nautical mile x
1 nautical mile
10
1 km
= 8262.50km
x
x
1000 m
3.28 ft
Distance = 74.290 x
a = 900 + 33052’ = 123052’
b = 900 - 140 35’ = 75025’
C = 151012’ - 120059’ = 30013’
b
cos c = cos a cos b + sin a sin b cos C
b
c
6. A Philippine Airlines Plane on one of its trip is to fly from Manila (Latitude 14035’N,
Longitude 120059’E) to Sydney, Australia (Latitude 33052’S, Longitude 151012’E).
Determine the distance in nautical miles from Manila to Sydney.
●
●
A
sin 750 22’
sin 74.290
sin 340 15’
=
=
sin 830 28’
sin A
sin B
●
A = 35.510
Equator
B = 86.950
Bearing of Moscow from Manila = N 35.510 W
South Pole
Bearing of Manila from Moscow = N 86.950 E
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