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

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- Jul 17th 2009, 01:03 PMtherealdrag0Circle 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 PMskeeter
- Jul 17th 2009, 01:25 PMtherealdrag0
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 PMskeeter
- Jul 17th 2009, 01:38 PMtherealdrag0
Ah; good move.

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

Thank you for the help. - Jul 17th 2009, 01:50 PMSoroban
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?

Code:`* * * A`

* *

* / | *

* r / | *

/ |

* / | *

* O * θ |d *

* \ | *

\ |

* r \ | *

* \ | *

* *

* * * B

You know: .$\displaystyle \begin{array}{ccc}d \:=\:\text{length of chord }AB \\

\theta \:=\: \text{in{c}luded angle }AOB \end{array}$

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

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

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

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