I got that NK = 12 and JK = 24 but i don't know the other two?
mhanifa:
What would be mNLK?
SmokeyBrown:
For arcs MK and JPK, we can start by calculating the circumference of the entire circle.
We know the radius of the circle from LK, 15. From that, we know that the circumference of the circle is 2*pi*15, or 30*pi.
Now, arc MK is a fraction of the total circumference, but what fraction is that? If we find the measure of angle L in the right triangle NLK (let's call that angle L), then the length of arc MK will be (L/360)*(30*pi).
This is because the total "angle" measure of a circle is 360, so that smaller angle would be a fraction of the total 360
mhanifa:
You have all three side measures of triangle NLK
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m4nt1c0r3:
So how would we find angle L?
SmokeyBrown:
First, to find the measure of angle L.
Based on the right triangle inside the circle, we could say that cos(L) = 9/15
From this, we can also calculate that
\[L = \cos^{-1} (9/15)\]
SmokeyBrown:
My calculations say that L would be about 53 degrees, based on the above
m4nt1c0r3:
That's 53.13, so would we use the formula \ ?
m4nt1c0r3:
s = r x theta*
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SmokeyBrown:
I'm sorry, I'm not familiar with that formula. What do you use it for?
m4nt1c0r3:
Arc length
mhanifa:
No, you just need angle measures
SmokeyBrown:
I think mhanifa is right, actually, and that was my misunderstanding.
The question asks for the angle measures of the arcs, not the lengths of the arcs. Looking more closely at the question, I see that the boxes for mJPK and mMK are meant to be in degrees, so we've already found mMK, like you said
mhanifa:
You found one for mMK
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ITSurBOY:
take a D and out the middle dot on each corner and then count the number you will get your answer
m4nt1c0r3:
Point K is 36.9 degrees what do we do with that?
mhanifa:
Please ignore point K
ITSurBOY:
@m4nt1c0r3 wrote:
Point K is 36.9 degrees what do we do with that?
now measure M
m4nt1c0r3:
oh ok
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SmokeyBrown:
This is fortunate for us, since we don't have to do quite as much work.
mJM happens to be symmetrical with mMK, across line ML; the angle measures are the same.
So, (though the question does not ask for it) we know mJM is 53.13, just as mMK is 53.13. That gives us a total of 106.26 degrees in those two arcs. And it just so happens that arc JPK spans the entire rest of the circle.
So, to find mJPK, we just have to calculate 360 - 106.26. Unless I am mistaken, of course
m4nt1c0r3:
@smokeybrown wrote:
This is fortunate for us, since we don't have to do quite as much work.
mJM happens to be symmetrical with mMK, across line ML; the angle measures are the same.
So, (though the question does not ask for it) we know mJM is 53.13, just as mMK is 53.13. That gives us a total of 106.26 degrees in those two arcs. And it just so happens that arc JPK spans the entire rest of the circle.
So, to find mJPK, we just have to calculate 360 - 106.26. Unless I am mistaken, of course
Would it be symmetrical only because it's perpendicular?
mhanifa:
SmokeyBrown you are correct
SmokeyBrown:
@m4nt1c0r3 wrote:
Would it be symmetrical only because it's perpendicular?
Maybe "symmetrical" wasn't very precise of me to say. I should have specified that JN and NK are the same length, and NL is shared between the two arcs. If we imagine another right triangle connecting JLN, it will be the same as the triangle we previously worked with, NLK.
So, the triangles are the same, and the angles are the same, meaning the arcs JM and MK have the same angle measures
m4nt1c0r3:
@smokeybrown wrote:
@m4nt1c0r3 wrote:
Would it be symmetrical only because it's perpendicular?
Maybe "symmetrical" wasn't very precise of me to say. I should have specified that JN and NK are the same length, and NL is shared between the two arcs. If we imagine another right triangle connecting JLN, it will be the same as the triangle we previously worked with, NLK.
So, the triangles are the same, and the angles are the same, meaning the arcs JM and MK have the same angle measures
Got it.
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m4nt1c0r3:
Yay we got it all right :D Thanks guys
mhanifa:
The perpendicular bisector of JK, that passes through the center makes JN = NK and respective arcs mJM = mJK