Thread: Circle From Included Angle and Chord Length

1. 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

2. 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.

3. 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.

4. 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?

5. Ah; good move.

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

Thank you for the help.

6. Hello, therealdrag0!

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

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 . . .