What is the equation for sine alpha when alpha is greater than 90 degrees?

[fawk3s]

[PLAIN]http://img266.imageshack.us/img266/8206/triangles.png

This is not a homework question, but it is something I can't wrap my mind around.
What is the equation for sine alpha where alpha is greater than 90 degrees?

For the first triangle its easy: sine alpha = b/c
But what is it for the second triangle?

Thanks in advance,
fawk3s

 

[tiny-tim]

Hi fawk3s!

(have an alpha: α )

hmm … why have you swapped a and b in the second diagram?

Go back to the first diagram, and just slide the vertical line (marked "b") to the left (dragging c with it), until the bottom line is length a again (but to the left) …

then b and c will be the same, and a will be minus the same, so sin is the same, and cos and tan are minus.

(it's easier to follow this if you draw a circle with the α angle at the centre and c as a radius)

 

[fawk3s]

Well, I think I follow. It seems quite easy and now when I think about it, it makes pretty good sense.
It didnt at first, I have to admit, since it seemed like you were finding the sin for the same angle as in the first diagram, but it all makes sense now because "a" is negative. Just too bad I didnt think of it myself :(

Thanks tiny-tim !