Spherical Law of Sines/Proof 2

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Theorem

Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.

Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.


Then:

$\dfrac {\sin a} {\sin A} = \dfrac {\sin b} {\sin B} = \dfrac {\sin c} {\sin C}$


Proof

Spherical-Cosine-Formula-2.png

Let $A$, $B$ and $C$ be the vertices of a spherical triangle on the surface of a sphere $S$.

By definition of a spherical triangle, $AB$, $BC$ and $AC$ are arcs of great circles on $S$.

By definition of a great circle, the center of each of these great circles is $O$.

Let $O$ be joined to each of $A$, $B$ and $C$.

Let $P$ be an arbitrary point on $OC$.

Construct $PQ$ perpendicular to $OA$ meeting $OA$ at $Q$.

Construct $PR$ perpendicular to $OB$ meeting $OB$ at $R$.

In the plane $OAB$:

construct $QS$ perpendicular to $OA$
construct $RS$ perpendicular to $OB$

where $S$ is the point where $QS$ and $RS$ intersect.

Let $OS$ and $PS$ be joined.

Let tangents be constructed at $A$ to the arcs of the great circles $AC$ and $AB$.

These tangents contain the spherical angle $A$.

But by construction, $QS$ and $QP$ are parallel to these tangents.

Hence $\angle PQS = \sphericalangle A$.

Similarly, $\angle PRS = \sphericalangle B$.

Also we have:

\(\ds \angle COB\) \(=\) \(\ds a\)
\(\ds \angle COA\) \(=\) \(\ds b\)
\(\ds \angle AOB\) \(=\) \(\ds c\)


It is to be proved that $PS$ is perpendicular to the plane $AOB$.

By construction, $OQ$ is perpendicular to both $PQ$ and $QS$.

Thus $OQ$ is perpendicular to the plane $PQS$.

Similarly, $OR$ is perpendicular to the plane $PRS$.

Thus $PS$ is perpendicular to both $OQ$ and $OR$.

Thus $PS$ is perpendicular to every line in the plane of $OQ$ and $OR$.

That is, $PS$ is perpendicular to the plane $OAB$.

In particular, $PS$ is perpendicular to $OS$, $SQ$ and $SR$

It follows that $\triangle PQS$ and $\triangle PRS$ are right triangles.


From the right triangles $\triangle OQP$ and $\triangle ORP$, we have:

\(\text {(1)}: \quad\) \(\ds PQ\) \(=\) \(\ds OP \sin b\)
\(\text {(2)}: \quad\) \(\ds PR\) \(=\) \(\ds OP \sin a\)
\(\text {(3)}: \quad\) \(\ds OQ\) \(=\) \(\ds OP \cos b\)
\(\text {(4)}: \quad\) \(\ds OR\) \(=\) \(\ds OP \cos a\)


From the right triangles $\triangle PQS$ and $\triangle PRS$, we have:

\(\ds PS\) \(=\) \(\ds PS \sin \angle PRS\)
\(\ds \) \(=\) \(\ds PQ \sin A\)
\(\ds PS\) \(=\) \(\ds PR \sin \angle PRS\)
\(\ds \) \(=\) \(\ds PR \sin B\)
\(\ds \leadsto \ \ \) \(\ds OP \sin b \sin A\) \(=\) \(\ds OP \sin a \sin B\) from $(1)$ and $(2)$
\(\ds \leadsto \ \ \) \(\ds \dfrac {\sin a} {\sin A}\) \(=\) \(\ds \dfrac {\sin b} {\sin B}\)


The result follows by applying this technique mutatis mutandis to the other angles of $ABC$.

$\blacksquare$


Sources

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