Chapter 5.4, Problem 98E

To determine

To derive: The formula of angle between two lines $\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

Answer to Problem 98E

  $\theta ={15}^{\circ }$

Explanation of Solution

Given: The equation of two lines in the given figure $y_1=m_{1}x+b_1\text{ and }{y}_2=m_{2}x+b_2.$

Assume that both lines have positive slopes.

  $y=x\text{ and }y=\frac{1}{\sqrt{3}}x$

Let the angle of line $y_1$ with positive direction of $x-$ axis be $\alpha .$

  $\therefore m_1=\tan \alpha$

Let the angle of line $y_2$ with positive direction of $x-$ axis be $\beta .$

  $\therefore m_2=\tan \beta$

According to given figure,

Let the angle of line $y_1$ with positive direction of $x-$ axis be $\alpha .$

  $\theta =\beta -\alpha$

Applying tan both sides

  $\begin{array}{c}\tan \theta =\tan \left(\beta -\alpha \right)\\ \tan \theta =\frac{\tan \beta -\tan \alpha }{1+\tan \beta \tan \alpha }\\ \tan \theta =\frac{m_2-m_1}{1+m_1{m}_2}\end{array}$

The given equation of line $y=x\text{ and }y=\frac{1}{\sqrt{3}}x$

  $\begin{array}{l}y=x\Rightarrow y=m_{1}x+b\\ m_1=1\\ y=\frac{1}{\sqrt{3}}x\Rightarrow y=m_{2}x+b\\ m_2=\frac{1}{\sqrt{3}}\end{array}$

Substituting the value $m_1\text{ and }{m}_2\text{ in formula}$

  $\begin{array}{c}\tan \theta =\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}\\ \tan \theta =\frac{\sqrt{3}-1}{\sqrt{3}+1}\\ \theta ={\tan }^{-1}\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)\\ \theta ={15}^{\circ }\end{array}$

Hence, the angle between the given line is ${15}^{\circ }$ .