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Line AB is tangent to the circle at B. m<A = 14 and mBC=112
a. Find x.
b. Find y.
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OpenSessame
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mathstudent55
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|dw:1377144471353:dw|
OpenSessame
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\[14=.5(112-X)\]
Is that how i would do it?
OpenSessame
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Or \[14=.5(x-112)\]
mathstudent55
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The first one.
OpenSessame
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Okay...So 84?
mathstudent55
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|dw:1377145019880:dw|
mathstudent55
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Yes
OpenSessame
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is 84 X? i feel like im missing something..
mathstudent55
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Yes, x = 84. Now you still need y.
OpenSessame
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i wouldnt use the same theorem right?
mathstudent55
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No. This a different case. You need a different theorem for this one.
mathstudent55
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y is the measure of an angle. It is not a length. Perimeter or circumference is a length.
OpenSessame
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Okay...So then I'm clueless.
mathstudent55
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An angle formed by a tangent and a chord that intersect on the circle is half the measure of the intercepted arc.
OpenSessame
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Right!
mathstudent55
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|dw:1377145667642:dw|
OpenSessame
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But we need to find the arc...right?
mathstudent55
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What is the degree measure of arc CD?
mathstudent55
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What is the degree measure of an entire circle as an arc?
OpenSessame
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360
mathstudent55
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Right.
Which parts of the circle below do you know the measures of?
|dw:1377145802550:dw|
OpenSessame
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SO 360-112-84
OpenSessame
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164?
mathstudent55
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|dw:1377145885955:dw|
OpenSessame
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So 164, right?
mathstudent55
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Correct.
Now you need half of that for y.
OpenSessame
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82!
mathstudent55
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Bingo!
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Thank you!
mathstudent55
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You deserve a medal.
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wlcm
OpenSessame
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I just need someone to guide me threw the first couple of times then i start getting it lol
mathstudent55
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That's the whole idea of studying and having a teacher or a tutor.
OpenSessame
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Do you know anything about probability distribution?And Im studying online, so i miss out on that just a little bit!
mathstudent55
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No, sorry, probability is not my thing.
OpenSessame
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Alright, thanks for your help!
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wlcm
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Thanks to you and @Directrix I understand geometry!
mathstudent55
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That's great, thanks. Keep asking questions. That's how you learn. You can tag me any time. If I'm not on, I'll see it the next time I'm on, and if you didn't get any response yet, I'll try to help.