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May 18th, 2014, 01:11 AM   #1
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Smile I need help.

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Guys I need help on this one and,,please ignore the fact that I circled A as the answer
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May 18th, 2014, 02:02 PM   #2
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C: 108

No trigonometry required, just calculating angles is sufficient.

Start with angle OKL.

Notice 3 isosceles triangles: MOK, JOK and JOM (all sides equal to radius).
So angleOKL = angleOMK, angleOKJ = angleOJK, angleOJM = angleOMJ.
Also evident that angle KJM = 180-28-46 = 106.
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May 18th, 2014, 07:20 PM   #3
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$\displaystyle \text{By the inscribed angle theorem }\angle{JOM}=2\angle{JKM}=56^{\circ}$

$\displaystyle \text{Hence }56+2(46+\angle{KMO})=180\text{ and }\angle{KMO}=\angle{MKO}=16^{\circ}$

$\displaystyle x=\angle{KLJ}=180-28-44=108^{\circ}$
Thanks from Denis
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May 19th, 2014, 07:43 AM   #4
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Thanks Greg; how do you remember all them theorems?

This problem can be done in 3 steps without using that theorem, agree?

x = KMO = MKO ; indirectly given that KJM=106
Step 1
MJO = 46+x, KJO = 28+x; so 46+x+28+x = 106 : x = 16

Step2
MOJ = 180 - 2(46+x) = 180 - 2(46+16) = 56

Etape numero trois:
MLO = 180 - x - MOJ = 180 - 16 - 56 = 108

Think the Habs can beat the Rangers?
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May 19th, 2014, 10:23 AM   #5
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Agree.

Price is out for the remainder of the conference final. Looks grim ...
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May 19th, 2014, 10:57 AM   #6
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Quote:
Originally Posted by greg1313 View Post
Price is out for the remainder of the conference final. Looks grim ...
We've seen miracles in the playoffs before:
perhaps Peter Budaj will become one...
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