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Thread: What about triangle?

  1. #1
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    What about triangle?

    What can say about triangle in which is

    $$ b=2c\cos {\alpha} $$

    Same question about triangle in which is

    $$ b=2c\sin \frac {\beta}{2} $$
    Last edited by ns1954; Dec 2nd 2018 at 03:39 AM.
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  2. #2
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    Re: What about triangle?

    Law of cosines

    $\displaystyle a^2-c^2=b(b-2 c \cos \alpha )$

    Therefore

    $\displaystyle b=2 c \cos \alpha $ if and only if $\displaystyle a=c$
    Thanks from topsquark
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  3. #3
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    Re: What about triangle?

    $$ b = 2c\sin \frac {\beta}{2} \Leftrightarrow \sin^2 \frac {\beta}{2} = \frac {b^2}{4c^2} $$

    and

    $$ 1- \cos \beta = 2 \sin^2 \frac {\beta}{2} = \frac {b^2}{2c^2} \ \ and \ \ \cos \beta = \frac {a^2 + c^2 - b^2}{2ac}$$

    gives equality

    $$ 1 - \frac {a^2 + c^2 - b^2}{2ac} = \frac {b^2}{2c^2} $$

    and then to

    $$ (c - a)(ac - c^2 + b^2) = 0 $$

    and if

    $$ ac - c^2 + b^2) \ne 0 $$

    which is missing detail, then $ a = c $

    Obviosly $ b < 2c \ ... $
    Last edited by ns1954; Dec 2nd 2018 at 12:12 PM.
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  4. #4
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    Re: What about triangle?

    consider triangle ABC

    $\displaystyle c=1$

    $\displaystyle a=\sqrt{3}-1$

    $\displaystyle b=\frac{-1+\sqrt{3}}{ \sqrt{2}}$

    $\displaystyle b=2c \sin \left(\frac{\beta }{2}\right)$

    but $\displaystyle a\neq c$
    Thanks from topsquark and ns1954
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  5. #5
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    Re: What about triangle?

    Awesome example. How you find it? BTW, problem is taken from book, so book is wrong.
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