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I am trying to solve this problem, but I am having trouble. So what's giving me trouble is that I am not given the units for the given arc lengths, so I cannot tell if it's 20 degrees or 20 pi radians. I know how to do this problem if it was in degrees but I am trying to do this problem for pi radians.

Anyways, I need to find arc JK here is my procedure:

I know the circumference of a circle is $C=2 \pi r$ which is $27 \pi=2 \pi r$ which means $r=13.5$

So to find angle mHQJ=$\frac{20}{13.5}=1.48$ pi radians and mKQM=$\frac{40}{13.5}=2.96 $ pi radians

Now I am going to add 2.96+1.48=4.4 pi radians and converting that to degrees I get 254 degrees which is where I am stuck because angle mHQM=$180$ degrees but the sum is greater than $180$ degrees.

Therefore, I am stuck. Any ideas on solving this problem?

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    $\begingroup$ $20 (\pi$ radians $) = 20 (180^0) = 3600^0$ $\endgroup$ – Mick Aug 14 '16 at 14:33
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    $\begingroup$ The only sensible interpretation of that problem is $\angle HQJ=20°$ and $\angle KQM=40°$. $\endgroup$ – Intelligenti pauca Aug 14 '16 at 14:40
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a. The arc measure of $JK$ is $120^{\circ}$.

The total arc length of $HM$ is $180^{\circ}$. Therefore, $JK=180^{\circ}-20^{\circ}-40^{\circ}=120^{\circ}$.

b. The length of arc $JK$ is $9\pi$.

The circumference is $27\pi$. Therefore, you can set up a proportion from what you know now.

$\frac{120^{\circ}}{360^{\circ}}=\frac{JK}{27\pi}$.

This is the part/whole strategy.

You get $JK=9\pi$.

c. HM is 27.

$HM=2r$, and you know that $r=13.5$. Therefore, $HM=27$.

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