# Thread: Angle Between Two Lines

1. ## 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)

2. Assuming that we're talking about lines in the 2D space:

$\displaystyle \tan \left( \theta \right) = \frac{{m_2 - m_1 }} {{1 + m_1 m_2 }}$

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

$\displaystyle \tan \left( \theta \right) = \frac{{\tan \left( {\theta _2 } \right) - \tan \left( {\theta _1 } \right)}} {{1 + \tan \left( {\theta _1 } \right)\tan \left( {\theta _1 } \right)}}$

using the following well known trigonometric identity:

$\displaystyle \tan \left( {\theta _2 \pm \theta _1 } \right) = \frac{{\tan \left( {\theta _2 } \right) \pm \tan \left( {\theta _1 } \right)}} {{1 \mp \tan \left( {\theta _1 } \right)\tan \left( {\theta _1 } \right)}}$

we get:

$\displaystyle \tan \left( \theta \right) = \tan \left( {\theta _2 - \theta _1 } \right)$

now let's look at the following diagram:

thus:

$\displaystyle \begin{gathered} \theta _1 + 180 - \theta _2 + \theta = 180 \hfill \\ \theta = \theta _2 - \theta _1 \hfill \\ \end{gathered}$

combined with the previous identity this proves the equality.

3. ## Great!

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

$\displaystyle \tan \left( \theta \right) = \frac{{m_2 - m_1 }} {{1 + m_1 m_2 }}$

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

$\displaystyle \tan \left( \theta \right) = \frac{{\tan \left( {\theta _2 } \right) - \tan \left( {\theta _1 } \right)}} {{1 + \tan \left( {\theta _1 } \right)\tan \left( {\theta _1 } \right)}}$

using the following well known trigonometric identity:

$\displaystyle \tan \left( {\theta _2 \pm \theta _1 } \right) = \frac{{\tan \left( {\theta _2 } \right) \pm \tan \left( {\theta _1 } \right)}} {{1 \mp \tan \left( {\theta _1 } \right)\tan \left( {\theta _1 } \right)}}$

we get:

$\displaystyle \tan \left( \theta \right) = \tan \left( {\theta _2 - \theta _1 } \right)$

now let's look at the following diagram:

thus:

$\displaystyle \begin{gathered} \theta _1 + 180 - \theta _2 + \theta = 180 \hfill \\ \theta = \theta _2 - \theta _1 \hfill \\ \end{gathered}$

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

### let theta be acute angle between two lines L1 and L2

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