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Math Help - calculate the angle

  1. #1
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    calculate the angle

    ok well when you look at the attachment you can see the diagram i am supposed to be using to figure out the answer to the questions. i dont know where to start i havent had geometry since high school, and to be honest we dont use much of it in calculus. so anyways im stuck on #16 and #15 on the handout. if arc HI = 80 degrees and arc JI = 75 what is the measure of angle HKI? see my first question is why did they break up the arcs? if you add them together wouldnt arc HJ = 155 degrees?IMG.pdf my next problem im having is that the center of the circle is Q but the center of the inscribed box is K, i found some identities but they are using the center of the circle as reference, which means i cant use them? any help that could be provided would be greatly appreciated..thanks in advance...
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  2. #2
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    Quote Originally Posted by slapmaxwell1 View Post
    ok well when you look at the attachment you can see the diagram i am supposed to be using to figure out the answer to the questions. i dont know where to start i havent had geometry since high school, and to be honest we dont use much of it in calculus. so anyways im stuck on #16 and #15 on the handout. if arc HI = 80 degrees and arc JI = 75 what is the measure of angle HKI? see my first question is why did they break up the arcs? if you add them together wouldnt arc HJ = 155 degrees?IMG.pdf my next problem im having is that the center of the circle is Q but the center of the inscribed box is K, i found some identities but they are using the center of the circle as reference, which means i cant use them? any help that could be provided would be greatly appreciated..thanks in advance...
    Hi slapmaxwell1,

    [16]

    \widehat{HG}=80

    \widehat{JI}=75

    As far as I can tell here, \widehat{HG} is not needed.

    Angle JHI is an inscribed angle = one-half of arc JI. \angle JHI=37.5^{\circ}

    \widehat{HG}=75


    Angle GIH is an inscribed angle = one-half of arc GH. \angle GIH =37.5^{\circ}

    This means that triangle HIK is isosceles. The vertex angle HKI that your are looking for = 180 - 75 = 105 degrees.

    [15]

    \widehat{HL}=40^{\circ}

    Calculate \widehat{IO}

    A diameter perpendicular to a chord bisects the chord and its subtended arc.

    That means \widehat{HL}=\widehat{LI}=40^{\circ}

    Now find \widehat{IO} by subtracting this from 180.

    Answer: 140^{\circ}
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  3. #3
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    ok you wouldnt happen to have the formulas? or some info i can read to help me through these problems? i mean in case you get tied up or busy? :O) i wish someone could go through all the math books and jus write down or summarize the formulas and explain how they work, that would be a very valuable book. thanks for your help, im going to print this out and study it right now.
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  4. #4
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    thx again for your help, i really appreciate you taking the time to answer my question. i did get ahold of a geometry book with a list of theorems? i wish it had pictures to better illustrate the properties...lol
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