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Math Help - triangle inscribed in a circle...

  1. #1
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    triangle inscribed in a circle...

    Triangle ABC is inscribed in a circle as shown in the figure. AX,BY and CZ bisect A,BAnd C respectively. If BAC= 56 degrees, ABC=64 degrees. Find the angles of XYZ. A,Y,C,X,Z in clockwise.
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  2. #2
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    Hello, snigdha!

    Triangle ABC is inscribed in a circle as shown in the figure.

    AX,BY, CZ bisect \angle A, \angle B, \angle C, respectively.

    If \angle BAC= 56^o,\;\angle ABC= 64^o,

    find the angles of \Delta XYZ\/\text{ and }\,\angle C
    Thereom: An inscribed angle is measured by one-half its intercepted arc.


    Since \angle BAC = 56^o, then: . \text{arc}(BXC) \,=\,112^o

    Since \angle ABC = 64^o, then: . \text{arc}(AYC) \,=\,128^o

    Hence: . \text{arc}(BZA) \;=\;360^o - 112^o - 128^o \;=\;120^o

    Therefore: . \boxed{\angle C \:=\:60^o}


    Since CZ bisects \angle C\!:\;\angle ACZ \,=\, \angle BCZ \,=\, 30^o \quad\Rightarrow\quad \text{arc}(BZ) \,=\,\text{arc}(ZA) \,=\, 60^o

    Since BY bisects \angle B\!:\;\angle ABY \,=\, \angle CBY \,=\,32^o \quad\Rightarrow\quad \text{arc}(AY) \,=\, \text{arc}(YC) \,=\,64^o

    Hence: . \text{arc}(ZAY) \:=\:60^o + 64^o \:=\:124^o

    . . Therefore: . \boxed{\angle X \:=\:62^o}


    \text{arc}(XBZ) \:=\:56^o + 60^o \:=\:116^o \quad\Rightarrow\quad \boxed{\angle Y \:=\:58^o}


    \text{arc}(XCY) \:=\:56^o + 64^o \:=\:120^o \quad\Rightarrow\quad\boxed{ \angle Z \:=\:60^o}

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