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Math Help - Double-angle Formulae

  1. #1
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    Double-angle Formulae

    Having a bit of trouble with this one. Can anyone help me out?

    Many thanks.

    Q.
    In the triangle pqr (see attachment), |\angle qrp|=90^{\circ} & |rp| = h. s is a point on [qr] such that |\angle spq|=2B & |\angle rps|=45^{\circ}-B, 0^{\circ}<B<45^{\circ}
    . If |sr|=h\tan(45-B), show that |qs|=2h \tan B.

    Attempt: \frac{|qr|}{h}=\tan(2B+45-B)
    |qr|=h\tan(45+B)
    |qr|-|sr|=h\tan(45+B)-h\tan(45-B)
    |qs|=\frac{h\tan45+h\tan B}{1-\tan45\tan B}-(\frac{h\tan45-h\tan B}{1+\tan45\tan B})
    |qs|=\frac{2h\tan B}{1-\tan^245\tan B}
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  2. #2
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    Re: Double-angle Formulae

    Hello, GrigOrig99!

    By the way, don't you know that \tan45^o = 1\,?


    Having a bit of trouble with this one.
    No wonder . . . the problem is stated incorrectly.

    \text{In }\Delta PQR,\,\angle QRP=90^o\,\text{ and }\,RP = h.\;S\text{ is a point on }QR
    \text{ such that }\angle SPQ= 2B\text{ and }\angle RPS =45^o-B,\;0^o<B<45^o
    \text{If }SR=a=h\tan(45-B),\,\text{ show that: }\:x=QS = 2h\tan ({\color{red}2}B).
    Code:
                                          o P
                                      * * |
                                  *2B *   |
                              *     * 45-B|
                          *       *       | h
                      *         *         |
                  *           *           |
              *             *             |
        Q o * * * * * * * o * * * * * * * o R
                  x       S       a
    Note that \angle QPR \:=\:2B + 45^o-B \:=\:45^o+B

    In \Delta QPR\!:\;\tan(45+B) \,=\,\frac{x+a}{h} \quad\Rightarrow\quad x+a \:=\:h\tan(45+B) .[1]

    In \Delta SPR\!:\;\tan(45-B) \,=\,\frac{a}{h} \quad\Rightarrow\quad a \:=\:h\tan(45-B) .[2]

    Subtract [1] - [2]:

    . . x \;=\;h\tan(45+b) - h\tan(45-B) \;=\;h\left[\tan(45+B) - \tan(45-B)\right]

    . . x \;=\;h\left[\frac{\tan45 + \tan B}{1-\tan45\tan B} - \frac{\tan45 - \tan B}{1 + \tan45\tan B}\right] \;=\;h\left[\frac{1+ \tan B}{1-\tan B} - \frac{1 - \tan B}{1+\tan B}\right]

    . . x \;=\;h\left[\frac{(1+\tan B)^2 - (1-\tan B)^2}{(1-\tan B)(1+\tan B)}\right] \;=\;h\left[\frac{1 + 2\tan B + \tan^2\!B - 1 + 2\tan B -\tan^2\!B}{1-\tan^2\!B}\right]

    . . x \;=\;h\left[\frac{\4\tan B}{1-\tan^2\!B}\right] \;=\;2h\left[\frac{2\tan B}{1-\tan^2\!B}\right] \;=\;2h\tan(2B)
    Last edited by Soroban; November 8th 2012 at 05:21 AM.
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  3. #3
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    Re: Double-angle Formulae

    Great. Thank you very much.
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