Trigonometry/For Enthusiasts/Triangles on a Sphere

Spherical Triangle

A spherical triangle is a part of the surface of a sphere bounded by arcs of three great circles. (For a discussion of great circles, see The Distance from New York to Tokyo.) Because the surface of a sphere is curved, the formulae for triangles do not work for spherical triangles. In particular, the sum of the three angles always exceeds $180^{\circ }$ or $\pi$ radians.

The amount in radians that the sum of the angles exceeds $\pi$ is known as the spherical excess of the triangle, and is proportional to the triangle's area. Thus, for a very small triangle, the excess is small and the sum of the angles is close to $\pi$ radians, reflecting the fact that a very small part of a sphere is not appreciably curved. This is why ordinary trigonometry, which assumes that you are working on a flat surface, is accurate enough for short distances on the Earth's surface.

The length of a side is usually expressed as the angle that the side subtends at the centre of the sphere, so sides as well as angles are expressed in degrees or radians and we can talk about the sine or cosine of a side as well as of an angle.

The Sine Theorem

This is similar to the sine theorem for ordinary triangles. If the sides are $a,b,c$ and the angles are $\alpha ,\beta ,\gamma$ then

${\frac {\sin(a)}{\sin(\alpha )}}={\frac {\sin(b)}{\sin(\beta )}}={\frac {\sin(c)}{\sin(\gamma )}}$

Exercise: Compare to sine formula for Triangles on a Plane For small triangles on a unit sphere $\sin(\alpha )\approx \alpha$ , provided we are measuring in radians, and we can make the percentage error as small as we like by taking a sufficiently small triangle. We therefore approach the sine formula for triangles on a plane as the triangles get very small. For a spherical triangle with three right angles, that's a large triangle taking up ${\tfrac {1}{8}}$ of the sphere's surface area, the sine formula for triangles on a plane is still correct even if applied to the spherical triangle! So, when is the plane triangle formula most in error if applied to a spherical triangle?

The Cosine Theorem

Despite its name, this does not look much like the cosine theorem for ordinary triangles. With the above notation,

$\cos(a)=\cos(b)\cos(c)+\sin(b)\sin(c)\cos(\alpha )$

In particular, if $\alpha$ is a right angle so $\cos(\alpha )=0$ , this formula becomes

$\cos(a)=\cos(b)\cos(c)$

This can be regarded as analogous to Pythagoras' theorem.

Compare to Pythagoras' Theorem for Triangles on a Plane For small triangles on a unit sphere $\cos(\alpha )\approx 1-{\frac {\alpha ^{2}}{2}}$ , again providing we are measuring in radians, and we can make the percentage error as small as we like by taking a sufficiently small triangle. Applying this approximation to the analogue of Pythagoras' Theorem, we get $1-{\frac {a^{2}}{2}}=\left(1-{\frac {b^{2}}{2}}\right)\left(1-{\frac {c^{2}}{2}}\right)\approx 1-{\frac {b^{2}}{2}}-{\frac {c^{2}}{2}}$ or $a^{2}=b^{2}+c^{2}$ We therefore approach Pythagoras' Theorem for triangles on a plane as the triangles get very small.

There are of course similar formulae involving $\beta$ or $\gamma$ instead of $\alpha$ . Thus, if two sides of a spherical triangle and the angle between them are known, we can find the third side from this formula. If all three sides are known, we can find all three angles from this and the similar formulae.

Half-angle formulae

Let $s={\frac {a+b+c}{2}}$ . Then

$\sin \left({\frac {\alpha }{2}}\right)={\sqrt {\frac {\sin(s-b)\sin(s-c)}{\sin(b)\sin(c)}}}$

$\cos \left({\frac {\alpha }{2}}\right)={\sqrt {\frac {\sin(s)\sin(s-a)}{\sin(b)\sin(c)}}}$

$\tan \left({\frac {\alpha }{2}}\right)={\sqrt {\frac {\sin(s-b)\sin(s-c)}{\sin(s)\sin(s-a)}}}$

with similar formulae for $\beta ,\gamma$ .

The Polar Triangle

If the corners of a spherical triangle are $ABC$ , the great circle of which side $BC$ is part divides the sphere into two hemispheres, each with a pole at its centre, in the same way that the equator divides the Earth into the northern and southern hemispheres, each with a pole. Let $A'$ be the pole in the hemisphere containing $A$ . Similarly, we can define $B'$ and $C'$ . $A'B'C'$ is the polar triangle of $ABC$ , and has sides $a',b',c'$ . Then

$A'=180^{\circ }-a\ ;\ B'=180^{\circ }-b\ ;\ C'=180^{\circ }-c$

$a'=180^{\circ }-A\ ;\ b'=180^{\circ }-B\ ;\ c'=180^{\circ }-C$

Note: We will need proofs of all these theorems.