# Spherical Law of Sines

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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 1

 $\ds \sin b \sin c \cos A$ $=$ $\ds \cos a - \cos b \cos c$ Spherical Law of Cosines $\ds \leadsto \ \$ $\ds \sin^2 b \sin^2 c \cos^2 A$ $=$ $\ds \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c$ $\ds \leadsto \ \$ $\ds \sin^2 b \sin^2 c \paren {1 - \sin^2 A}$ $=$ $\ds \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c$ Sum of Squares of Sine and Cosine $\ds \leadsto \ \$ $\ds \sin^2 b \sin^2 c - \sin^2 b \sin^2 c \sin^2 A$ $=$ $\ds \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c$ multiplying out $\ds \leadsto \ \$ $\ds \paren {1 - \cos^2 b} \paren {1 - \cos^2 c} - \sin^2 b \sin^2 c \sin^2 A$ $=$ $\ds \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c$ Sum of Squares of Sine and Cosine $\ds \leadsto \ \$ $\ds 1 - \cos^2 b - \cos^2 c + \cos^2 b \cos^2 c - \sin^2 b \sin^2 c \sin^2 A$ $=$ $\ds \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c$ multiplying out $\text {(1)}: \quad$ $\ds \leadsto \ \$ $\ds \sin^2 b \sin^2 c \sin^2 A$ $=$ $\ds 1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c$ rearranging and simplifying

Let $X \in \R_{>0}$ such that:

$X^2 \sin^2 a \sin^2 b \sin^2 c = 1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c$

Then from $(1)$:

 $\ds \dfrac {X^2 \sin^2 a \sin^2 b \sin^2 c} {\sin^2 b \sin^2 c \sin^2 A}$ $=$ $\ds \dfrac {1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c} {1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c}$ $\ds \leadsto \ \$ $\ds X^2$ $=$ $\ds \dfrac {\sin^2 A} {\sin^2 a}$

In a spherical triangle, all of the sides are less than $\pi$ radians.

The same applies to the angles.

$\sin \theta > 0$ for all $0 < \theta < \pi$

Hence the negative root of $\dfrac {\sin^2 A} {\sin^2 a}$ does not apply, and so:

$X = \dfrac {\sin A} {\sin a}$

Similarly, from applying the Spherical Law of Cosines to $\cos B$ and $\cos C$:

 $\ds \sin a \sin c \cos B$ $=$ $\ds \cos b - \cos a \cos c$ $\ds \sin a \sin b \cos C$ $=$ $\ds \cos c - \cos a \cos b$

we arrive at the same point:

 $\ds X$ $=$ $\ds \dfrac {\sin B} {\sin b}$ $\ds$ $=$ $\ds \dfrac {\sin A} {\sin a}$

where:

$X^2 \sin^2 a \sin^2 b \sin^2 c = 1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c$

as before.

Hence we have:

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

$\blacksquare$

## Proof 2 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 $ABC$.

$\blacksquare$

## Also known as

Some sources refer to this as just the sine-formula.

## Historical Note

The Spherical Law of Sines was first stated by Abu'l-Wafa Al-Buzjani.

It was rediscovered by Regiomontanus, and published in his De Triangulis Omnimodus of $1464$.

## Sources

... where he misattributes it to Georg Joachim Rhaeticus