Using **arc 1(A)** &** arc 2(B)** and difference in the two arc radius is **W**, Find** r1, r2** ?

Guest Aug 27, 2017

#1**+1 **

This diagram has a lot of information, and it can be somewhat difficult to decipher. With 2 radii, it should be possible to figure out the central angle. Before we start, we must understand arc length and its formula. The formula is below

\(\text{Arclength}=\frac{m^{\circ}}{360^{\circ}}*2\pi r\)

m = measure of the central angle, in degrees

r = radius of the circle

First, let's create 2 equations that demonstrate the relationship of r_{1} in this diagram:

\(l=\frac{m^{\circ}}{360^{\circ}}*2\pi r\) | I am using "l" as my chosen variable for arc length. Of course, we already know the arc length; it is A, or 32. And we know that arc length corresponds with r_{1}. |

\(32=\frac{m^{\circ}}{360^{\circ}}*2\pi r_1\) | Great! Now, I will solve for r_{1} now. First, let's simplify the right hand side of the equation. |

\(\frac{m^{\circ}}{360^{\circ}}*\frac{2\pi r_1}{1}\) | Multiply the fractions together. |

\(\frac{2\pi r_1 m^{\circ}}{360^{\circ}}\) | The numerator and denominator have a GCF of 2, so factor that out. |

\(\frac{\pi r_1 m^{\circ}}{180^{\circ}}\) | Reinsert this into the orginal equation. |

\(32=\frac{\pi r_1 m^{\circ}}{180^{\circ}}\) | Multiply by 180 on both sides of the equation. |

\(5760^{\circ}=\pi r_1 m^{\circ}\) | Divide by \(\pi m^{\circ}\) to isolate r_{1} |

\(r_1=\frac{5760^{\circ}}{\pi m^{\circ}}\) | |

Let's do the exact same process for r_{2}.

\(34=\frac{m^{\circ}}{360^{\circ}}*2\pi r_2\) | We already from before what the right hand side simplifies to, so let's do that. |

\(34=\frac{\pi r_2 m^{\circ}}{180^{\circ}}\) | Do the same process to solve for r_{2}. Multiply both sides by 180. |

\(6120^{\circ}=\pi r_2 m^{\circ}\) | Divide both sides by \(\pi m^{\circ}\). |

\(r_2=\frac{6120^{\circ}}{\pi m^{\circ}}\) | |

We have two equations, and they are the following:

{

\(r_2=\frac{6120^{\circ}}{\pi m^{\circ}}\)

\(r_1=\frac{5760^{\circ}}{\pi m^{\circ}}\)

{

Now, let's subtract the 2 equations from each other. Let's see what happens. Unfortuntaely, I am no genius with LaTeX, so I could not line up the equal signs. Hopefully, you get the point...

\(\begin{eqnarray*} r_2=\frac{6120^{\circ}}{\pi m^{\circ}}\\ -\left(r_1=\frac{5760^{\circ}}{\pi m^{\circ}}\right)\\ \end{eqnarray*}\)

Let's subtract the two equations together. Let's start with the left hand side subtraction. \(r_2-r_1=1.5\), according to the given info. \(\frac{6120^{\circ}}{\pi m^{\circ}}-\frac{5760}{\pi m^{\circ}}=\frac{360}{\pi m^{\circ}}\). Knowing this, we can now solve for the central angle.

\(1.5=\frac{360}{\pi m^{\circ}}\) | Multiply by \(\pi m^{\circ}\) on both sides. |

\(1.5\pi m^{\circ}={360}\) | Divide by \(1.5\pi\) |

\(m^{\circ}=\frac{360}{1.5\pi}\) | We can simplify the right hand side further, actually. First, let's convert 1.5 into a fraction. |

\(\frac{360}{1.5\pi}=\frac{360}{\frac{3\pi}{2}}\) | In a fraction, \(\frac{a}{\frac{b}{c}}=\frac{a*c}{b}\). Let's apply it. |

\(\frac{360}{\frac{3\pi}{2}}=\frac{360*2}{3\pi}\) | In the numerator and denominator, 3 can be factored out. |

\(m^{\circ}=\frac{120*2}{\pi}=\frac{240}{\pi}\) | |

Now that I know the central angle in its exact form, I can now substitute it back into the arc length formula to ge the measure of the radii. Let's do that! FIrst, I'll solve for r_{1} first.

\(l=\frac{m^{\circ}}{360^{\circ}}*2\pi r\) | As a reminder, this is the formula for arc length. Substitute the values we already know. |

\(32=\frac{m^{\circ}}{360}*2\pi r_1\) | I have decided to leave out the measure of the angle for now, so the equation does not look like a monstrosity. Multiply both sides by 360. |

\(11520=2\pi r_1m^{\circ}\) | Divide by 2 on both sides. |

\(5760=\pi r_1m^{\circ}\) | Now, let's plug in m. |

\(5760=\pi r_1\left(\frac{240}{\pi}\right)\) | Eliminate the pi's as they cancel out. |

\(5760=240r_1\) | Divide by 240 on both sides. |

\(r_1=\frac{5760}{240}=24units\) | I happen to know that 24^2=576, so this division was relatively easy for me. I will not forget the units, either! |

We don't have to do this same process for r_{2} luckily as \(r_1+1.5=r_2\). Therefore, \(r_2=24+1.5=25.5units\)

TheXSquaredFactor Aug 27, 2017

#2**+1 **

Note that the formula for the arc length, S, is

S = r * theta where "theta" is the central angle forming the arc and r is the radius

And let r = "r_{1}"

So....we have this system

32 = r * theta → 32/r = theta (1) and letting r_{2} = r + 1.5

34 = (r + 1.5) * theta (2)

Sub (1) into (2) for "theta" and we have that

34 = (r + 1.5) (32/r) multiply through by r

34r = (r + 1.5) (32) simplify

34r = 32r + 48 subtract 32r from both sides

2r = 48 divide both sides by 2

r = 24 = " r_{1} "

And

r + 1.5 = 24 + 1.5 = 25.5 = "r_{2}"

CPhill Aug 27, 2017