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Math Help - Triangle

  1. #1
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    Triangle

    What is the value of angle A and C in degrees ?
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  2. #2
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    Quote Originally Posted by totalnewbie
    What is the value of angle A and C in degrees ?
    Hello,

    because AB = BC then AB = BD and BC = BD because the Points A, C, D are points of the semicircle of Thales. B is the centre of this circle.

    Triangle ADB is an isosceles triangle, therefore the angle at A has a value of 30.

    The angle at D is a right one. Therefore the angle at C has a value of 60.

    Hope that this helps a little bit.

    Greetings

    EB
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  3. #3
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    I like the problem
    I solved it with some trigonometry.
    By law of sines we have,
    \frac{\sin A}{BD}=\frac{\sin 30^o}{AB} (1)
    \frac{\sin C}{BD}=\frac{\sin 60^o}{BC} (2)
    Invert (2) to get,
    \frac{BD}{\sin C}=\frac{BC}{\sin 60^o}
    Multiply (1) and (2) to get, remember AB=BC
    Thus,
    \frac{\sin A}{\sin C}=\frac{\sin 30^o}{\sin 60^o}
    Thus,
    \frac{\sin A}{\sin C}=\frac{\sqrt{3}}{3}
    But,
    \sin C=\cos (90^o-C)=\cos A
    Thus,
    \frac{\sin A}{\cos A}=\frac{\sqrt{3}}{3}
    Thus,
    \tan A=\frac{\sqrt{3}}{3}
    Thus,
    A=60^o
    Thus,
    C=30^o

    \mathbb{Q}.\mathbb{E}.\mathbb{D}
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  4. #4
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    would this work? i extended some sides and introduced parallel lines to derive the angles. not a very practical way. excuse my sloppy handwriting.
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    Last edited by c_323_h; March 22nd 2006 at 08:47 AM.
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