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Double-angle Formulae

  1. Nov 7, 2012 #1
    In the final proof, I have [itex]\frac{h-h\tan B}{\tan B}[/itex], rather than [itex]\frac{h-h\tan B}{1+\tan B}[/itex]. Can anyone help me out?

    Many thanks.

    1. The problem statement, all variables and given/known data

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

    2. Relevant equations

    3. The attempt at a solution

    [itex]h\tan(45-B)=h(\frac{\tan45-\tan B}{1+ tan45\cdot\tan B})=\frac{h-h\tan B}{1+\tan B}[/itex]

    Apply the sine rule: [itex]\frac{\sin(45-B)}{|sr|}=\frac{\sin(180-(90+45-B))}{h}[/itex]
    [itex]h(\sin(45-B))=|sr|\sin(45+B)[/itex]
    [itex]h(\sin\cos B-\cos45\sin B)=|sr|(\sin45\cos B+\cos45\sin B)[/itex]
    [itex]h(\frac{\cos B}{\sqrt{2}}-frac{sin B}{\sqrt{2}})=|sr|(\frac{\cos B}{\sqrt{2}}+\frac{\sin B}{\sqrt{2}})[/itex]
    [itex]|sr|=(\frac{h\cos B-h\sin B}{\sqrt{2}})/(\frac{\cos B}{\sin B}{\sqrt{2}})[/itex]
    [itex]\frac{h\sqrt{2}(\cos B- \sin B)}{\sqrt{2}(\cos B+\sin B})[/itex]
    [itex]h(\frac{\sin B}{\cos B}+\frac{\cos B}{\sin B})[/itex]
    [itex]h(-\tan B+\frac{1}{\tan B})[/itex]
    [itex]\frac{h-h\tan B}{\tan B}[/itex]
     

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  3. Nov 7, 2012 #2

    LCKurtz

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    What am I missing? I see right triangle srp with$$
    \tan(45-B) = \frac {sr} h$$Just solve for ##sr##.
     
  4. Nov 7, 2012 #3
    Looks like I was over-thinking this. Ok, thank you.
     
  5. Nov 7, 2012 #4

    SammyS

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    [itex]\displaystyle \frac{\cos B- \sin B}{\cos B+\sin B}\ne\frac{-\sin B}{\cos B}+\frac{\cos B}{\sin B}[/itex]

    You can't break up a fraction in that manner!

    attachment.php?attachmentid=52742&d=1352309813.jpg
     
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