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Math Help - Nature of the triangle

  1. #1
    Super Member dhiab's Avatar
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    Nature of the triangle

    Determine the nature of the triangle ABC, the measurements of his angles of which are respectively, a, b, c such as:
    \frac{{\sin a}}{{\sin b}} = 2\cos c<br />
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  2. #2
    MHF Contributor
    Grandad's Avatar
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    Hello dhiab
    Quote Originally Posted by dhiab View Post
    Determine the nature of the triangle ABC, the measurements of his angles of which are respectively, a, b, c such as:
    \frac{{\sin a}}{{\sin b}} = 2\cos c<br />
    It is isosceles, with \angle B = \angle C.

    This may be proved by trigonometry formulae:
    \frac{\sin A}{\sin B}=2\cos C

    \Rightarrow \frac{\sin(B+C)}{\sin B}= 2\cos C, since A = 180^o -(B+C)

    \Rightarrow \frac{\sin B \cos C + \cos B \sin C}{\sin B}=2\cos C

    \Rightarrow \cos C  + \cot B \sin C = 2\cos C

    \Rightarrow \cot B\sin C = \cos C

    \Rightarrow \tan C = \tan B

    \Rightarrow C = B, since B and C both lie between 0^o and 180^o
    or, using the Sine Rule, which gives:
    \frac{\sin A}{\sin B}=\frac{a}{b}=2\cos C

    \Rightarrow \cos C = \frac{\frac12a}{b}
    and then considering the attached diagram:
    P is the foot of the perpendicular from A to BC. Then
    \cos C = \frac{CP}{AC}= \frac{\frac12a}{b}

    \Rightarrow CP = \tfrac12a

    \Rightarrow P is the mid-point of CB

    \Rightarrow \triangle ABC is isosceles with AC = AB.
    Grandad
    Attached Thumbnails Attached Thumbnails Nature of the triangle-untitled.jpg  
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  3. #3
    MHF Contributor red_dog's Avatar
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    Another approach:

    \frac{\sin A}{\sin B}=2\cos C\Rightarrow\sin A=2\sin B\cos C\Rightarrow

    \Rightarrow \sin A=\sin(B+C)+\sin(B-C)\Rightarrow

    \Rightarrow\sin A=\sin A+\sin(B-C)\Rightarrow\sin(B-C)=0\Rightarrow B=C
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