I have seen this physical proof of the sine rule attributed to one English engineer named Stevinus (Stevens; cjeris informs me, in addition to the Latin name, that the diagram appears on his tombstone), sometime in the 16th century. I have no idea if this is indeed true. But the proof itself is delightfully simple and elegant; it appeals to our physical intuition to prove the mathematical fact. As such, its mathematical standing might be regarded as suspect. But even so, it remains illuminating.

Consider a triangle. Rest the triangle on a pedestal supporting one of its sides (c in glorious ASCIIvision below):

C
/`-.
a / `-. b
/ `-.
/ `-.
/________________:.
B c A

We wish to show that a/

sin(A) = b/sin(B). We shall show this using only a simple

prop!

Take a long loop of heavy chain. We assume that the links of the chain are very small and all weigh the same. Place it on the triangle:

@@@@
@/`-.@@@
a @/ `-.@@@ b
@/ `-.@@@
@/ `-.@@
@/________________:.@
@ c @
@ @
@ @
@ @
@ @
@ @
@ @
@ @
@ @
@ @
@@ @@
@@@@@

Observe!

*The chain stays still!* That was the bit where we used

Physics; we can toss it back into the

cupboard now. (If it rotated in one direction, it would continue to

accelerate, and we'd have an accelerating

perpetuum mobile -- clearly impossible

(at least for so simple a machine...).) This means that the

forces exerted to rotate the chain

clockwise (links on b, and on the right half of the loose bit below c) are equal to the forces exerted to

rotate the chain

counterclockwise (links on a, and on the left half of the loose bit below c).

The shape adopted by a chain supported at the two points A and B is entirely symmetrical. So the weight of the chain below side c is evenly distributed between the left and right halves. Thus, the force exerted by the chain along b is equal to the force exerted by the chain along a. But the force exerted by a chain of length l at an angle of x is proportional to l*sin(x). So we have that

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

-- exactly the sine law!