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Math Help - another circle arc

  1. #1
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    Unhappy another circle arc

    I thought maybe I was doing this right but now I don't think so. Can someone help me out with this and the others I am still not understanding?
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  2. #2
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    Quote Originally Posted by sanee66 View Post
    I thought maybe I was doing this right but now I don't think so. Can someone help me out with this and the others I am still not understanding?
    Hello,

    I assume that CE is a tangent to the circle.

    If so \bigtriangleup (ABC) is an isosceles triangle with base AB.

    Let M be the centre of the circle and CE is tangent to the circle:
    1. (CM) \bot (CE)
    2. (CE) \parallel (AB)
    3. (CM) \bot (AB)
    4. (CM) is perpendicular bisector of (AB) thus \bigtriangleup (ABC) is an isosceles triangle.
    5. thus \angle(CBA) = \angle(BAC)

    6. \angle(DCA) is an exterior angle.

    7. \angle(CBA) + \angle(BAC) = \angle(DCA)
    8. Therefore: \angle(CBA) = \angle(BAC) = 55^\circ

    9. Therefore \angle(BMC) = 2 \cdot \angle(BAC) = 110^\circ

    10.The arc(BC) belongs to the angle \angle(BMC). You only have to convert the measure in degree into radians:

    arc(BC) = \frac{110^\circ}{360^\circ} \cdot 2 \pi = \frac{11}{18} \pi \approx 1.92

    Maybe you should seek for the tangent-chord-theorem(?) [I don't know if this is the appropriate name in English] which states that the angle between the chord (here BC, name of angle \angle(BCF)) is as large as the angle at the circumference (here \angle(BAC)).
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  3. #3
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    Quote Originally Posted by earboth View Post
    Hello,

    I assume that CE is a tangent to the circle.

    If so \bigtriangleup (ABC) is an isosceles triangle with base AB.

    Let M be the centre of the circle and CE is tangent to the circle:
    1. (CM) \bot (CE)
    2. (CE) \parallel (AB)
    3. (CM) \bot (AB)
    4. (CM) is perpendicular bisector of (AB) thus \bigtriangleup (ABC) is an isosceles triangle.
    5. thus \angle(CBA) = \angle(BAC)

    6. \angle(DCA) is an exterior angle.

    7. \angle(CBA) + \angle(BAC) = \angle(DCA)
    8. Therefore: \angle(CBA) = \angle(BAC) = 55^\circ

    9. Therefore \angle(BMC) = 2 \cdot \angle(BAC) = 110^\circ

    10.The arc(BC) belongs to the angle \angle(BMC). You only have to convert the measure in degree into radians:

    arc(BC) = \frac{110^\circ}{360^\circ} \cdot 2 \pi = \frac{11}{18} \pi \approx 1.92

    Maybe you should seek for the tangent-chord-theorem(?) [I don't know if this is the appropriate name in English] which states that the angle between the chord (here BC, name of angle \angle(BCF)) is as large as the angle at the circumference (here \angle(BAC)).
    Optionally

    \measuredangle~BAC=\measuredangle~ACE=\measuredang  le~ECD=55^\circ\implies\stackrel{\displaystyle\fro  wn}{BC}=110^\circ
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  4. #4
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    Hello, sanee66!

    Another approach . . .


    We are told that \angle ACD = 110^o and CE bisects that angle.
    . . Hence: . \angle ACE = 55^o

    Since AB \parallel CE,\;\angle BAC \:= \:\angle ACE \:=\:55^o .(alt-int. angles)

    \angle BAC is an inscribed angle, measured by one-half its intercept arc.

    . . Therefore: . \text{arc }BC \:=\:110^o

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