Colegio San Agustin-BacolodCollege of Engineering

Sperical Trigonometry

In partial fulfilment of the requirements in Math 402E: Plane and Spherical Trigonometry

Jefferson P. Aries BSME1October 2015

SPHERICAL TRIGONOMETRY

- the study of the relations connecting the sides and angles of a spherical triangle- is a branch of trigonometry that concerns with triangles extracted from the surface of

the sphere- basic in astronomy, surveying and navigation- time of day, directions of sailing and flying, positions of ships, airplanes and reference

points

Great circle – a circle cut from a sphere by a plane through the center of the sphere. It divides the sphere into two equal parts.Ex: circles NWS and P’BAP

Arc length – angle subtended by the arc at the center of the sphereEx: arc AB has arc length of θ

Quarter of a great circle has an arc length of 90ºAn arc of 1’ on a sphere equal to the earth in volume is a nautical mile and has 6080.3 ft.Thus 90º has 5400 nautical miles.

Poles of the circle – a line through the center O of a sphere perpendicular to the plane of a circle on the sphere cuts the sphere in two pointsEx: P and P’

Polar distance of a circle – least distance on a sphere from a point on the circle to its poleEx: Polar distance of great circle is 90º.

Polar distance of TQR is arc PR.

Theorem: If the planes of two great circles are perpendicular, then their great circle passes through the poles of the other and conversely.

Lune – one of the parts in which a sphere is divided by two diametral planesEx: PABP’SR

Earth – ellipsoidal in shape about 7918 statute miles in diameter

Axis – line through the center where the earth revolves

Poles – points in which its axis cuts the surface; north and south poles

Equator – is a great circle cut from the earth by a plane perpendicular to the axis of the earth through the center

Parallel of latitude (parallel) – is a small circle cut from the earth by a plane parallel to the equatorial plane

Meridian – is half of a great circle on the earth terminated by the north pole and the south pole

Latitude – angular distance from the equator along a meridian through the place- measured north or south of the equator from 0º-90º

Longitude – angle at either pole between the meridian passing through the point and some fixed meridian known as the prime meridian- measured east or west of the PM from 0º-180º

SPHERICAL TRIANGLE- consists of three arcs of great circles that form the boundaries of a portion of a spherical surface- the vertices will be denoted by A, B and C, and the sides opposite by a, b and c

Important Propositions from Solid Geometry:1. If two angles of a spherical triangle are equal, the sides opposite are equal; and

conversely.2. If two angles of a spherical triangle are unequal, the sides opposite are unequal, and the

greater side lies opposite the greater angle; and conversely.3. The sum of two sides of a spherical triangle is greater than the third side. In symbols,

a + b > c4. The sum of the sides of a spherical triangle is less than 360º. In symbols,

0º < a + b + c < 360º5. The sum of the angles of a spherical triangle is greater than 180º and less than 540º;

that is,180º < A + B + C < 540º

6. The sum of any two angles of a spherical triangle is less than 180º plus the third angle; that is,

A + B < 180º + C

SOLUTION OF SPHERICAL TRIANGLES

Law of cosines for the sides:cos a = cos b cos c + sin b sin c cos Acos b = cos a cos c + sin a sin c cos Bcos 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 acos B = - cos A cos C + sin A sin C cos bcos C = - cos A cos B + sin A sin B cos c

Law of sines:sin Asina

= sinBsinb

= sinCsin c

RIGHT SPHERICAL TRIANGLE- one of the angles is 90º

Formulas Relating to Right Spherical Triangles:• tan a = tan c cos B• tan a = sin b tan A• tan b = sin a tan B• tan b = tan c cos A• cos c = cot A cot B• sin a = sin c sin A• sin b = sin c sin B• cos c = cos a cos b• cos A = cos a sin B• cos B = cos b sin A

NAPIER’S RULES

co-A = complement of A = 90º - Aco-B = complement of B = 90º - Bco-c = complement

a, b, co-c, co-A and co-B are called the circular parts

If a is the middle part, b and co-B are adjacent to a and co-c and co-A are opposite to a.

Recall: (same with co-B and co-c)sin A = cos co-A = cos (90º - A)cos A = sin co-A = sin (90º - A)tan A = cot co-A = cot (90º - A)cot A = tan co-A = tan (90º - A)sec A = csc co-A = csc (90º - A)csc A = sec co-A = sec (90º - A)

Napier’s Rule:

I. The sine of any middle part is equal to the product of the cosines of the opposite parts.

II. The sine of any middle part is equal to the product of the tangents of the adjacent parts.

sin middle = cos opposite = tan adjacent

LQ1: In a right spherical triangle an oblique angle and the side opposite are of the same quadrant.

LQ2: When the hypotenuse of a right spherical triangle is(a) less than 90º, the two legs are of the same quadrant and conversely.(b) greater than 90º, one leg is of the first quadrant, the other of the second, and conversely.

Quadrantal Triangles- a spherical triangle having a side equal to 90º

CLASSICAL METHODS OF SOLVING SPHERICAL TRIANGLES

Case 1: Two sides and included angle are given.Case 2: Two angles and the included side.Case 3: Two sides and an angle opposite one of them.Case 4: Two angles and a side opposite one of them.Case 5: Three sides are given.Case 6: Three angles are given.

NAPIER’S ANALOGIES:

TERRESTRIAL SPHERE:

where: E = spherical excess in degrees E = ( A+B+C) - 180

1 . sin

12

(A−B )

sin12

(A+B )=

tan12

(a−b )

tan12c

2 . cos

12

( A−B )

cos12

( A+B )=

tan12

(a+b )

tan12c

3 . sin

12

(a−b )

sin12

(a+b )=

tan12

( A−B )

cot12C

4 . cos

12

(a−b )

cos12

(a+b )=

tan12

(A+B )

cot12C

Area of sphere =π r2E180

Problems:

1. Two places A and B on the earth have the follow latitudes and longitudes:A(40◦N, 18◦E) and B (0◦N, 58◦E). Find the angle of departure from A to B of the great circle route.

Given:

Required: ∠NAB

Solution:

By the given latitudes and longitudes, AN = 50◦, BN = 90◦ and ∠ANB = 58◦ − 18◦ = 40◦.Using the cosine formula,

Then, by the sine formula:

2.

Solution:

3.

Solution:

4. Find side a given that b  =  7.5m., c  =  5.9m., A  =  49o

Solution:

Cos(a)  =  [Cos(b) . Cos(c)] +[Sin(b) . Sin(c) .Cos(A)]

Substituting the values given above, we have:

cos a=[cos(7.5)cos (5.9)]+[sin(7.5) .sin (5.9).cos (49o)]= [0.991 x 0.995]  +  [0.13 x 0.10 x 0.66]

=  0.986  +  0.009

a=cos−1 0.995

a = 5.73m.

5.

Solution: