# Thread: Triangle inside of circle on graph - angle problem

1. ## Triangle inside of circle on graph - angle problem

The question wants me to show how ∠BOC = 2(∠BAC)

Here is the graph
http://i686.photobucket.com/albums/v...ied__/Math.jpg

(sorry if it isn't put together all that well, I just whipped it up using the "Graph" program and "Paint")

I'm a good math student, but this just has me stuck.
The unit we are working on right now is basically the Sin Law, Cosine Law, angles, solving similar triangles, etc.

2. Oh, I'd also like to mention that I'm aware of the circle law or something about how when you have a angle going through the middle of the circle and another one going on it's circumference the angle in the middle is double the size.
My teacher wants us to solve and show.
But I really don't know how to solve... LOL - yikes.

3. Well, there's the straight-forward way. You got yerself two triangles, you know the coordinates of the vertices, therefore you know the side-lengths, therefore you can calculate each angle.

Looks like that's not the intended method, though. I'm betting the sine/cosine laws will be what you use, but there's a chance that one of the equations sin(2x) = 2sinx cosx, sin(x/2) = +/- squareroot((1-cosx)/2), cos(2x) = (cosx)^2 + (sinx)^2 will be used, if you've talked about these already.

I'll think more about it, have to go for now.

4. Okay, that's got me thinking. I'll work on it tomorrow in class some more and see what I can come up with, then come back here.
Thanks.

5. Originally Posted by wunnymush13
The question wants me to show how ∠BOC = 2(∠BAC)

(sorry if it isn't put together all that well, I just whipped it up using the "Graph" program and "Paint")

$10000 x^2 + 114200 x + 10000 y^2 - 42800 y = 1143$