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

This post continues the series of navigation related exercises which were requested by navigation and maths teachers. When the post expires, its contents will be included under the menu heading ‘Navigation Exercises’ on the home page.

**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 = 32^{o}

Reminder. The cosine rule states that a^{2} = b^{2} + c^{2} – 2bc Cos.(A)

**Suggested Solution:**

a^{2} = b^{2} + c^{2} – 2bc Cos.(A)

substituting the above values gives: a^{2} = 73 – 48 Cos.(32^{o})

= 32.3 cm

therefore, a = 5.68 cm

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

To revise spherical trigonometry click here.

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 = 49^{o}

**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(49^{o})]

= [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 90^{o} – the declination,

CZ represents the latitude of the observer,

PZ represents 90^{o} – 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. = 45^{o} North, Dec. = 18^{o} North, and Time = 3 hours after noon, find the distance of the observer from the Sun’s GP.

**Suggested Solution:**

EX = 18^{o }therefore PX = 90^{o} – 18^{o} = 72^{o}

CZ = 45^{o }therefore PZ = 90^{o} – 45^{o} = 45^{o}

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

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.(45^{o}) x Cos.(72^{o})] + [Sin (45^{o}) x Sin(72^{o}) x Cos.(45^{o})]

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’.

Return to the blog www.astronavigationdemystified.com

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

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.

Thanks this really helps, more samples pls 🙂