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Math Help - Angle Between Two Lines

  1. #1
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    Angle Between Two Lines

    Let t = theta for short

    Let L1 and L2 denote two nonvertical intersecting lines, and let theta t denote the acute angle between L1 and L2.

    Show that tan(t) = (m2 - m1)/[1 + (m1)(m2)], where m1 and m2 are the slopes of L1 and L2, respectively.

    HINT GIVEN BY TEACHER:

    Use the facts that tan(t1) = m1 and tan(t2) = m2)
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  2. #2
    Senior Member Peritus's Avatar
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    Assuming that we're talking about lines in the 2D space:

    <br />
\tan \left( \theta  \right) = \frac{{m_2  - m_1 }}<br />
{{1 + m_1 m_2 }}<br />

    now using the fact that the slope of the line equals the tangent of the angle between the line and the positive x axis:

    <br />
\tan \left( \theta  \right) = \frac{{\tan \left( {\theta _2 } \right) - \tan \left( {\theta _1 } \right)}}<br />
{{1 + \tan \left( {\theta _1 } \right)\tan \left( {\theta _1 } \right)}}

    using the following well known trigonometric identity:

    <br />
\tan \left( {\theta _2  \pm \theta _1 } \right) = \frac{{\tan \left( {\theta _2 } \right) \pm \tan \left( {\theta _1 } \right)}}<br />
{{1 \mp \tan \left( {\theta _1 } \right)\tan \left( {\theta _1 } \right)}}

    we get:

    <br />
\tan \left( \theta  \right) = \tan \left( {\theta _2  - \theta _1 } \right)

    now let's look at the following diagram:


    thus:

    <br />
\begin{gathered}<br />
  \theta _1  + 180 - \theta _2  + \theta  = 180 \hfill \\<br />
  \theta  = \theta _2  - \theta _1  \hfill \\ <br />
\end{gathered} <br />

    combined with the previous identity this proves the equality.
    Last edited by Peritus; August 28th 2008 at 04:02 AM.
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  3. #3
    MHF Contributor
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    Great!

    Quote Originally Posted by Peritus View Post
    Assuming that we're talking about lines in the 2D space:

    <br />
\tan \left( \theta \right) = \frac{{m_2 - m_1 }}<br />
{{1 + m_1 m_2 }}<br />

    now using the fact that the slope of the line equals the tangent of the angle between the line and the positive x axis:

    <br />
\tan \left( \theta \right) = \frac{{\tan \left( {\theta _2 } \right) - \tan \left( {\theta _1 } \right)}}<br />
{{1 + \tan \left( {\theta _1 } \right)\tan \left( {\theta _1 } \right)}}

    using the following well known trigonometric identity:

    <br />
\tan \left( {\theta _2 \pm \theta _1 } \right) = \frac{{\tan \left( {\theta _2 } \right) \pm \tan \left( {\theta _1 } \right)}}<br />
{{1 \mp \tan \left( {\theta _1 } \right)\tan \left( {\theta _1 } \right)}}

    we get:

    <br />
\tan \left( \theta \right) = \tan \left( {\theta _2 - \theta _1 } \right)

    now let's look at the following diagram:


    thus:

    <br />
\begin{gathered}<br />
\theta _1 + 180 - \theta _2 + \theta = 180 \hfill \\<br />
\theta = \theta _2 - \theta _1 \hfill \\ <br />
\end{gathered} <br />

    combined with the previous identity this proves the equality.
    Exactly what I needed!!
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