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Math Help - help with angle and figure

  1. #1
    Member helloying's Avatar
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    help with angle and figure

    pls help me solve part(a). thank you.
    Attached Thumbnails Attached Thumbnails help with angle and figure-img017.jpg  
    Last edited by helloying; June 28th 2009 at 06:20 AM.
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  2. #2
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    This should have been posted in the Trigonometry subforum.

    \begin{aligned}<br />
\tan(45^{\circ} + \alpha) &= p \\<br />
\frac{\tan 45^{\circ} + \tan \alpha}{1 - \tan 45^{\circ}\tan \alpha} &= p \\<br />
\frac{1 + \tan \alpha}{1 - \tan \alpha} &= p \\<br />
1 + \tan \alpha &= p(1 - \tan \alpha) \\<br />
1 + \tan \alpha &= p - (\tan \alpha)p \\<br />
\end{aligned}
    \begin{aligned}<br />
(\tan \alpha)p + \tan \alpha&= p - 1 \\<br />
(\tan \alpha)(p + 1) &= p - 1 \\<br />
\tan \alpha &= \frac{p - 1}{p + 1}<br />
\end{aligned}

    What you have scribbled on the top is actually correct, just not simplified:
    \frac{1 - p}{-1 - p} = \frac{-(p - 1)}{-(p + 1)} = \frac{p - 1}{p + 1}

    01
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  3. #3
    Member helloying's Avatar
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    Hi actually i meant the 1st part as in part (a). but thanks for telling me the simplied ans for the part(i).do u know how to solve part (a)?
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  4. #4
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    Quote Originally Posted by helloying View Post
    Hi actually i meant the 1st part as in part (a). but thanks for telling me the simplied ans for the part(i).do u know how to solve part (a)?
    Using AB and BC you can find AD as
    AD = ABcosθ + BCsinθ
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