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

 

 

Guest Aug 27, 2017
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2+0 Answers

 #1
avatar+1088 
+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 r1 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 r1.
\(32=\frac{m^{\circ}}{360^{\circ}}*2\pi r_1\) Great! Now, I will solve for r1 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 r1
\(r_1=\frac{5760^{\circ}}{\pi m^{\circ}}\)  
   

 

Let's do the exact same process for r2

 

\(34=\frac{m^{\circ}}{360^{\circ}}*2\pi r_2\) 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 r2. 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 r1 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 r2 luckily as \(r_1+1.5=r_2\). Therefore, \(r_2=24+1.5=25.5units\)

TheXSquaredFactor  Aug 27, 2017
 #2
avatar+76069 
+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  = "r1"        

 

So....we have this system

 

32  = r * theta     →   32/r  = theta    (1)        and  letting r2 = 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      =   " r1 "

 

And

 

r + 1.5  =  24 + 1.5  = 25.5  =  "r2"

 

 

 

cool cool cool

CPhill  Aug 27, 2017
edited by CPhill  Aug 27, 2017
edited by CPhill  Aug 27, 2017

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