Civil Engineering May 2022
DEFINITION OF TERMS:
1. Great Circle - circle on the surface of a sphere,
whose plane passes through the center of the
sphere.
Review Innovations
-The sum of the angles of a spherical triangle is
greater than 1800 and less than 5400.
Spherical Trigonometry
OBLIQUE SPHERICAL TRIANGLE:
- Has none of its angles equal to 900
Law of Sine
sin a sin b sin c
=
=
sin A sin B sin C
RIGHT SPHERICAL TRIANGLE:
is one with right angle.
2. Small Circle - circle constructed by a plane
crossing the sphere not in its center.
Law of Cosine 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
3. Spherical Angle - an angle formed by the
intersection of two great circles.
4. Spherical Triangle - a triangle on the surface of
the sphere formed by the intersection of three great
circles.
5. Polar Distance – least distance on a sphere from a
point on the circle to its pole.
6. Latitude – angular distance of the point from the
equator ranges from 00 to 900 at the poles.
7. Longitude – angular distance between the prime
meridian and the meridian through the point
ranges from 00 at the prime meridian to 1800
eastward and -1800 westward.
PROPERTIES OF SPHERICAL TRIANGLE:
Spherical Excess,
E = A + B + C - 1800
Spherical Defect,
D = 3600 – (a + b + c)
- The greater side has the greater angle opposite to it.
- The sum on any two sides is greater than the third
side.
-The sum of the sides of a spherical triangle is less
than 3600.
- The sum of any two angles is less than 1800 plus the
third angle.
Manila FB: @ReviewInnovationsOfficial
 (02) 8735-9161
0919-227-9194
Law of Cosine 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 = complement of
Note:
sin co-A = cos A
cos co-A = sin A
tan co-A = cot A
AREA OF SPHERICAL TRIANGLE:
Napier’s Rule:
Rule 1: (Sin-Ta-Ad Rule)
The sine of any middle part is equal to the product
of the tangents of its adjacent parts.
πr 2 E
A=
1800
Rule 2: (Sin-Co-Op Rule)
The sine of any middle part is equal to the product
of the cosines of its opposite parts.
tan
Important Rules:
1. In a right spherical triangle, an oblique angle and
the side opposite are of the same quadrant.
2. When the hypotenuse of a right spherical triangle
is less than 900, the two legs are of the same
quadrant and conversely.
3. When the hypotenuse of a right spherical triangle
is greater than 900, one leg is of the first quadrant
and the other of the second and conversely.
r = radius/radius of sphere
E = spherical excess
= sum of angles - 1800
= A + B + C - 1800
1
1
1
1
1
= √tan s tan (s − a) tan (s − b) tan (s − c)
4
2
2
2
2
s = (a + b + c)/2
Note:
Bi-rectangular spherical triangle
- 2 angles are right angles
Tri-rectangular spherical triangle
3 right angles
Terrestrial Sphere Problems:
Note:
1 minute of arc = 1 nautical mile
1 nautical mile = 6080 ft
1 statute mile = 5280 ft
1 nautical mile = 1.1516 statute mile
1 knot = 1 nautical mile per hour
Davao FB: Review Innovations Davao Branch
 (082) 221-1121 0930-256-0998
Civil Engineering May 2022
Sample Problems:
1. Solve the unknown angles and side of the
spherical triangle whose given parts are:
a = 720 27’
b = 610 49’
C = 900
2. Solve the triangle whose given parts are:
A = B = 640 37’, and b = 810 14’
3. Given the parts of spherical triangle, solve for
side C:
A = 1100 33’, C = 1400 48’ and a = 1520 19’
A. 18.280
C. 161.720
0
B. 20.19
D. 159.810
4. Find the area of the spherical triangle whose
angles are A = 600, B = 800, and C = 1040. The radius
of the sphere is 20 cm.
5. A plane left Manila (140 36’N, 1210 5’E) and flew
in the direction of S 320 E. At what longitude will it
cross the equator?
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.
7. Find the time it would take an airplane flying at
a speed of 1200 kph 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.
Manila FB: @ReviewInnovationsOfficial
 (02) 8735-9161
0919-227-9194
Review Innovations
Spherical Trigonometry
Problems for Practice:
1. A spherical triangle ABC has an angle C = 900 and
sides a = 500 and c = 800. Find the angle of side b in
degrees.
Answer: 74.330
2. Solve the spherical triangle whose given parts
are:
b = 72038’;
A = 1150 51’;
and C = 900
Answer: a = 1160 55’, c = 970 46’, B = 740 25’
3. Solve the spherical triangle whose given parts
are:
A = B = C = 1200
Answer: a = b = c = 109028.3’
4. In a spherical triangle ABC, A = 1160, B = 550 and
C = 800. Find the value of “a” in degrees.
Answer: a = 114.830
An airplane flew from Manila (140 36’ N, 1210 05’ E)
on a course S 300 W and maintaining a uniform
altitude.
5. At what point will the plane cross the equator?
6. What will be its course at that point?
Answer: (5) 1120 48’ E; (6) S 280 56’ W
An airplane flew from Manila (140 36’ N, 1210 05’ E)
at an average speed of 600 kph on a course S 32 0 E.
7. How long will it take to reach the equator?
8. At what point will it cross the equator?
9. What is the course of the airplane at the equator?
Answer: (8) 3.17 hours; (9) 1300 2’ E;
(10) S 300 51’ E
Davao FB: Review Innovations Davao Branch
 (082) 221-1121 0930-256-0998