# Positional Astronomy:

Spherical trigonometry

A great-circle arc, on the sphere, is the analogue of a straight line, on the plane.

Where two such arcs intersect, we can define the **spherical angle**

*either* as angle between the tangents to the two arcs, at the point of intersection,

*or* as the angle between the planes of the two great circles where they intersect at the centre of the sphere.

(Spherical angle is only defined where arcs of *great *circles meet.)

A **spherical triangle** is made up of three arcs of great circles, all less than 180°.

The sum of the angles is not fixed, but will always be greater than 180°.

If any side of the triangle is exactly 90°, the triangle is called **quadrantal**.

There are many formulae relating the sides and angles of a spherical triangle.

In this course we use only two: the *sine rule *and the *cosine rule*.

Consider a triangle ABC on the surface of a sphere with radius = 1.

*(See note) *

We use the capital letters A, B, C to denote the angles at these corners;

we use the lower-case letters a, b, c to denote the *opposite *sides.

(Remember that, in spherical geometry, the side of a triangle is the arc of a great circle,

so it is also an angle.)

Turn the sphere so that A is at the “north pole”,

and let arc AB define the “prime meridian”.

Set up a system of rectangular axes OXYZ:

O is at the centre of the sphere;

OZ passes through A;

OX passes through arc AB (or the extension of it);

OY is perpendicular to both.

Find the coordinates of C in this system:

x = sin(b) cos(A)

y = sin(b) sin(A)

z = cos(b)

Now create a new set of axes,

keeping the y-axis fixed and moving the “pole” from A to B

(i.e. rotating the x,y-plane through angle c).

The new coordinates of C are

x’ = sin(a) cos(180-B) = – sin(a) cos(B)

y’ = sin(a) sin(180-B) = sin(a) sin(B)

z’ = cos(a)

The relation between the old and new systems

is simply a rotation of the x,z-axes through angle c:

x’ = x cos(c) – z sin(c)

y’ = y

z’ = x sin(c) + z cos(c)

That is:

– sin(a) cos(B) = sin(b) cos(A) cos(c) – cos(b) sin(c)

sin(a) sin(B) = sin(b) sin(A)

cos(a) = sin(b) cos(A) sin(c) + cos(b) cos(c)

These three equations give us the formulae for solving spherical triangles.

The first equation is the **transposed cosine rule**,

which is sometimes useful but need not be memorised.

The second equation gives the **sine rule**. Rearrange as:

sin(a)/sin(A) = sin(b)/sin(B)

Similarly,

sin(b)/sin(B) = sin(c)/sin(C), etc.

So the sine rule is usually expressed as:

** sin(a)/sin(A) = sin(b)/sin(B) = sin(c)/sin(C) **

The third equation gives the **cosine rule**:

**cos(a) = cos(b) cos(c) + sin(b) sin(c) cos(A)
**and similarly:

**cos(b) = cos(c) cos(a) + sin(c) sin(a) cos(B)**

cos(c) = cos(a) cos(b) + sin(a) sin(b) cos(C)

cos(c) = cos(a) cos(b) + sin(a) sin(b) cos(C)

Here they are together:

**sine rule:**

** sin(a)/sin(A) = sin(b)/sin(B) = sin(c)/sin(C)**

**cosine rule:
cos(a) = cos(b) cos(c) + sin(b) sin(c) cos(A)
cos(b) = cos(c) cos(a) + sin(c) sin(a) cos(B)
cos(c) = cos(a) cos(b) + sin(a) sin(b) cos(C)**

The cosine rule will solve almost any triangle if it is applied often enough.

The sine rule is simpler to remember but not always applicable.

**Note **that both formulae can suffer from ambiguity:

*E.g.* if the sine rule yields

sin(x) = 0.5,

then x may be 30° *or* 150°.

Or, if the cosine rule yields

cos(x) = 0.5,

then x may be 60° *or *300° (-60°).

In this case, there is no ambiguity if x is a *side *of the triangle, as it must be less than 180°,

but there could still be uncertainty if an *angle *of the triangle was positive or negative.

So, when applying either formula, check to see if the answer is sensible.

If in doubt, recalculate using the other formula, as a check.

**Exercise:**

Alderney, in the Channel Islands, has longitude 2°W, latitude 50°N.

Winnipeg, in Canada, has longitude 97°W, latitude 50°N.

How far apart are they, in nautical miles, along a great-circle arc?

If you set off from Alderney on a great-circle route to Winnipeg,

in what direction (towards what azimuth) would you head?

Click here for the answer.

**Reference**

http://star-www.st-and.ac.uk/~fv/webnotes/chapter2.htm

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