Example 10 from Algebra and Trigonometry, 9.2 Sum and Difference Identities

Using Sum and Difference Formulas to Solve an Application Problem

Let $L_{1}$ and $L_{2}$ denote two non-vertical intersecting lines, and let $θ$ denote the acute angle between $L_{1}$ and $L_{2} .$ See Figure 7. Show that

\tan θ = \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}}

where $m_{1}$ and $m_{2}$ are the slopes of $L_{1}$ and $L_{2}$ respectively. (Hint: Use the fact that $\tan θ_{1} = m_{1}$ and $\tan θ_{2} = m_{2} .$ )

Diagram of two non-vertical intersecting lines L1 and L2 also intersecting the x-axis. The acute angle formed by the intersection of L1 and L2 is theta. The acute angle formed by L2 and the x-axis is theta 1, and the acute angle formed by the x-axis and L1 is theta 2.

Figure 7

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
Using the difference formula for tangent, this problem does not seem as daunting as it might.

$\begin{matrix} \tan θ & = & \tan (θ_{2} - θ_{1}) \\ = & \frac{\tan θ_{2} - \tan θ_{1}}{1 + \tan θ_{1} \tan θ_{2}} \\ = & \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}} \end{matrix}$

How did it go?

Solution

Solution