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Thread: 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 $\displaystyle \bigtriangleup (ABC)$ is an isosceles triangle with base $\displaystyle AB$.

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

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

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

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

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

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

    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 $\displaystyle \angle(BCF)$) is as large as the angle at the circumference (here $\displaystyle \angle(BAC)$).
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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 $\displaystyle \bigtriangleup (ABC)$ is an isosceles triangle with base $\displaystyle AB$.

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

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

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

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

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

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

    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 $\displaystyle \angle(BCF)$) is as large as the angle at the circumference (here $\displaystyle \angle(BAC)$).
    Optionally

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

    Another approach . . .


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

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

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

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

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