# Finding arc length of a sector.

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? • $20 (\pi$ radians $) = 20 (180^0) = 3600^0$ – Mick Aug 14 '16 at 14:33
• The only sensible interpretation of that problem is $\angle HQJ=20°$ and $\angle KQM=40°$. – Intelligenti pauca Aug 14 '16 at 14:40

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, $$HM=27$$.