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Math Help - Compound Angles

  1. #1
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    Compound Angles

    My answer is nearly right, except for the sign. Can anyone help me out here?

    Many thanks.

    Q.
    The diagram (see attachment) shows a triangle of height h metres. The angles A & B are such that A+B=45^{\circ}. By using the expansion of \tan (A+B), or otherwise, find the value of h.

    Attempt: \tan 45^{\circ}=1
    \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}=1
    \frac{(h/2)+(h/3)}{1-(h/2)(h/3)}=1
    \frac{h}{2}+\frac{h}{3}=1-\frac{h^2}{6}
    \frac{h^2}{6}+\frac{h}{2}+\frac{h}{3}-1=0
    h^2+3h+2h-6=0
    h^2+5h-6=0
    (h-1)(h+6)=0
    h = 1 or -6

    Ans.: (From text book): h = 6
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  2. #2
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    Re: Compound Angles

    Tan A= 2/h.
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  3. #3
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    Re: Compound Angles

    Thank you.
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