# Circle From Included Angle and Chord Length

• Jul 17th 2009, 01:03 PM
therealdrag0
Circle From Included Angle and Chord Length
I have the direct length between the start and end of a chord, and I have the included angle. How can I find the radius of the circle the arc is on?
Thanks
• Jul 17th 2009, 01:09 PM
skeeter
Quote:

Originally Posted by therealdrag0
I have the direct length between the start and end of a chord, and I have the included angle. How can I find the radius of the circle the arc is on?
Thanks

you can find the radius using the law of cosines.
• Jul 17th 2009, 01:25 PM
therealdrag0
Hm. I've looked at the page on Wikipedia: c^2 = a^2 + b^2 - 2*a*b*cos(y)

(because a=b) -> c^2 = 2a^2 - 2*a^2*cos(y)

but...I only have y (the angle) and c (the chord length), so I'm not sure what to do from here.
• Jul 17th 2009, 01:29 PM
skeeter
Quote:

Originally Posted by therealdrag0
Hm. I've looked at the page on Wikipedia: c^2 = a^2 + b^2 - 2*a*b*cos(y)

(because a=b) -> c^2 = 2a^2 - 2*a^2*cos(y)

but...I only have y (the angle) and c (the chord length), so I'm not sure what to do from here.

factor out $a^2$ from the right side ...

$c^2 = a^2(2 - 2\cos{y})$

can you solve for $a$ now?
• Jul 17th 2009, 01:38 PM
therealdrag0
Ah; good move.

$\sqrt{c^2 / (2 - 2\cos{y})} = a$

Thank you for the help.
• Jul 17th 2009, 01:50 PM
Soroban
Hello, therealdrag0!

Quote:

I have the length of a chord, and I have the included angle.
How can I find the radius of the circle?

Did you follow skeeter's suggestion?

Code:

              * * *    A           *          *         *          / | *       *        r /  |  *                   /  |       *          /    |  *       *      O * θ  |d  *       *          \    |  *                   \  |       *        r \  |  *         *          \ | *           *          *               * * *    B

You know: . $\begin{array}{ccc}d \:=\:\text{length of chord }AB \\
\theta \:=\: \text{in{c}luded angle }AOB \end{array}$

Law of Cosines: . $d^2 \:=\:r^2 + r^2 - 2r^2\cos\theta$

. . $d^2 \:=\: 2r^2 - 2r^2\cos\theta \:=\: 2r^2(1 - \cos\theta) \:=\:4r^2\left(\frac{1-\cos\theta}{2}\right) \quad\Rightarrow\quad d^2\:=\:4r^2\sin^2\!\left(\tfrac{\theta}{2}\right)$

Hence, we have: . $d \:=\:2r\sin\tfrac{\theta}{2} \quad\Rightarrow\quad r \:=\:\frac{d}{2\sin\frac{\theta}{2}} \quad\Rightarrow\quad r\;=\;\tfrac{1}{2}d\csc\tfrac{\theta}{2}$

There! . . . We just invented a formula for this problem . . .