Spherical Law of Sines/Proof 2
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
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
- 1976: W.M. Smart: Textbook on Spherical Astronomy (6th ed.) ... (previous) ... (next): Chapter $\text I$. Spherical Trigonometry: $9$. Alternative proofs of the formulae $\mathbf A$, $\mathbf B$ and $\mathbf C$.