The diagram shows a nine-point pegboard. We'll take the board radius as one (in other words it's a unit circle).
two isosceles triangles on a pegboard

Looking at triangle

OAM

, why is

AM

equal to

\sin 40^{\circ} ?

Looking at triangle $OAB$ , why is $AB$ equal to $2 \sin 20^{\circ}$ ?

Looking at triangle $ABM$ , why is $AM$ also equal to $AB \cos 20^{\circ}$ which works out as $2 \sin 20^{\circ}$ $\cos 20^{\circ}$

So it looks like the same $AM$ length value can be calculated by using $\sin 40^{\circ}$

and also calculated by using $2 \sin 20^{\circ} \cos 20^{\circ}$

Full Screen Version

Now switch to a

10

point pegboard and find the two ways to calculate the

AM

length on that board.

How about on a $12$ point board ? What general result is emerging ?
Draw a diagram and use it to provide reasoning which accounts for that general result.

Many thanks to Geoff Faux who introduced us to the merits of the 9 pin circular geo-board.
The boards, moulded in crystal clear ABS that can be used on an OHP (185 cm in diameter), together with a teacher's guide, are available from Geoff at Education Initiatives