Chapter 5.4, Problem 94E

To determine

To find:The angle between the given pair of lines $y=x$ and $y=\frac{1}{\sqrt{3}}x$ .

Answer to Problem 94E

The angle between the given pair of lines $y=x$ and $y=\frac{1}{\sqrt{3}}x$ is $15°$ .

Explanation of Solution

Given information:

The given two lines are $y=x$ and $y=\frac{1}{\sqrt{3}}x$ .

Calculation:

Let the graph for the two lines is $y_1=m_{1}x+b_1$ and $y_2=m_{2}x+b_2$ .

  

Let the angle made by $y_1=m_{1}x+b_1$ with the $x$ -axis be ${\theta }_1$ and $y_2=m_{2}x+b_2$ with the $x$ -axisbe ${\theta }_2$ .

  $\tan {\theta }_1=m_1,\tan {\theta }_2=m_2$ and $\theta ={\theta }_1-{\theta }_2$

Take tangent on both sides in equation $\theta ={\theta }_1-{\theta }_2$ .

  $\begin{array}{l}\tan \theta =\tan \left({\theta }_1-{\theta }_2\right)\\ \tan \theta =\frac{\tan {\theta }_1-\tan {\theta }_2}{1+\tan {\theta }_1\tan {\theta }_2}\\ \tan \theta =\frac{m_1-m_2}{1+m_1{m}_2}\text{}\end{array}$

Th above equation is the formulae for the angle between two lines with slopes $m_1$ and $m_2$ , $\left(m_1<m_2\right)$ .

Given lines are $y=x$ with slope $1$ , and $y=\frac{1}{\sqrt{3}}x$ with slope $\frac{1}{\sqrt{3}}$ .

  $m_1=1$ and $m_2=\frac{1}{\sqrt{3}}\left(\text{since}{m}_1>m_2\right)$

Calculate the angle.

  $\begin{array}{c}\tan \theta =\frac{m_1-m_2}{1+m_1{m}_2}\text{}\\ \tan \theta =\frac{1-\frac{1}{\sqrt{3}}}{1+\left(1\right)\frac{1}{\sqrt{3}}}\text{}=\frac{\sqrt{3}-1}{\sqrt{3}+1}\\ \tan \theta =\tan 15°\\ \theta =15°\end{array}$

Therefore, the angle between the given pair of lines $y=x$ and $y=\frac{1}{\sqrt{3}}x$ is $15°$ .