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Thread: Unable to get a step in the problem :(

  1. #1
    Senior Member
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    Exclamation Unable to get a step in the problem :(

    Hey guys



    The sides of triangle ABC are divided into segments of lengths a, b and x by the inscribed circle of centre O and radius r as shown. Find x in terms of a, b and r.

    The first part of the answer by the teacher is as shown:

    Let $\displaystyle \angle OAB = \alpha$ and $\displaystyle \angle OBA = \beta$
    Therefore, $\displaystyle \angle OCA = 90 (\alpha + \beta)$

    So $\displaystyle tan (90 (\alpha + \beta)) = \frac{x}{r}$
    .

    Could someone please explain to me how $\displaystyle tan (90 (\alpha + \beta)) = \frac{x}{r}$? For some reason, all I seem to be able to get is $\displaystyle tan (90 (\alpha + \beta)) = \frac{x}{r}$.

    Thanx.
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  2. #2
    MHF Contributor red_dog's Avatar
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    O is the intersection of the bisectors of the angles of triangle.

    Then $\displaystyle \widehat{A}=2\widehat{OAB}=2\alpha, \ \widehat{B}=2\widehat{OBA}=2\beta$

    $\displaystyle \widehat{OCA}=\frac{1}{2}\widehat{C}=\frac{1}{2}(1 80-2(\alpha+\beta))=90-(\alpha+\beta)$
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