# Exercise 2 = Application of Spherical Trigonometry

Exercise 2 – The application of spherical trigonometry in the solution of navigational problems.

Objective.  To exercise the application of the cosine rule in the solution of  spherical triangles.

Part 1.  Revising the application the cosine rule in the solutions of plane triangles. Task: Use the cosine rule to find side a in the triangle shown in the diagram above given that:  b =  8 cm,  c =  3 cm,  A =  32o

Reminder. The cosine rule states that a2 = b2 + c2 – 2bc Cos.(A)

Suggested Solution:

a2       =  b2 + c2 – 2bc Cos.(A)

substituting the above values gives:  a2   =  73 – 48 Cos.(32o)

=  32.3 cm

therefore,  a  =  5.68 cm

Part 2.  Applying the cosine rule in the solution of a spherical triangle.

As explained in the link above, we amend the cosine rule when we are working with spherical triangles.  For example, to find side a in the spherical triangle shown in the diagram below we have: Cos(a)  =  [Cos(b) . Cos(c)] +[Sin(b) . Sin(c) .Cos(A)] Example.  Find side a given that b  =  7.5m., c  =  5.9m., A  =  49o

Suggested 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

=  0.995

therefore a = 5.73m.

Part 3.  Applying spherical trigonometry in the solution of a navigational problem. In the diagram above,

Angle ZPX represents the time-angle,

P represents the North Pole,

Z represents the position of an observer on the Earth’s surface,

X represents the geographical position of the Sun (GP),

XZ represents the angular distance of the observer from the GP of the Sun,

EX represents the declination of the Sun,

PX represents 90o – the declination,

CZ represents the latitude of the observer,

PZ represents 90o – latitude,

E  represents a point on the Equator that lies on the same meridian of longitude as point X,

C represents a point on the Equator that lies on the same meridian of longitude as point Z.

Task:  Given the following: Lat. = 45o North, Dec. = 18o North, and Time = 3 hours after noon, find the distance of the observer from the Sun’s GP.

Suggested Solution:

EX  =  18o  therefore PX  =  90o – 18o  =  72o

CZ  =  45o  therefore PZ  =  90o – 45o  =  45o

Angle ZPX  =  45o (since 3 hours  =  a time angle of 3 x 15o).

Reminder.  As stated above, the formula for finding the third side of a spherical triangle, when the other two sides and their included angle is known, is:

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

Applying this formula to the diagram above, we have:

Cos.(XZ) =  [Cos.(PZ) x Cos.(PX)]  +  [Sin(PZ) x Sin(PX) x Cos.(ZPX)]

Substituting the values above gives:

Cos.(XZ) =  [Cos.(45o) x Cos.(72o)]  +  [Sin (45o) x Sin(72o) x Cos.(45o)]

Cos.(XZ) =  [0.707 x 0.309]  +  [0.707 x 0.951 x 0.707]

Cos.(ZX) =  0.218 + 0.475  =  0.693

therefore ZX   =  46.13  =  2767.8′

therefore distance of observer from GP = 2767.8 n.m.  (since 1′ of arc on the earth’s surface = approx. 1 nautical mile).

A fuller explanation of the application of spherical trigonometry in astro navigation is given in the book ‘Astro Navigation Demystified’.

### 3 Responses to Exercise 2 = Application of Spherical Trigonometry

1. Dale says:

Part 2 example problem: Can you mix linear quantities (meters) and arc degrees at will? I’d think not. Am I missing something? Thanks.

• Jack Case says:

Several people have had difficulty with this so I have explained it in my new post ‘The Relationship Between Altitude And Zenith Distance’
Hope this helps.

2. Joshua Bryan Monte Gatchalian says:

Thanks this really helps, more samples pls 🙂