Chapter 5.4, Problem 98E

To determine

Tofind:The angle between given pair of lines.

Explanation of Solution

Given information:

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

Formulaused:

The tangent of the angle a line makes with x-axis is called the slope of that line.The slope of straight line $y=mx+c$ is m.

Calculation:

Suppose two lines $y_1=m_{1}x+b_1$ and $y_2=m_{2}x+b_2$ are two lines making angles ${\theta }_1$ and ${\theta }_2$ with x-axis. Then by definition of the slopes of line $\tan {\theta }_1=m_1$ and $\tan {\theta }_2=m_2$ . Further, suppose the angle between lines is $\theta$ as shown in Figure-1 here.

  

Angle ${\theta }_1$ is exterior angle, therefore,

  ${\theta }_1={\theta }_2+\theta$ Or $\theta ={\theta }_1-{\theta }_2$.........(1)

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

  $\tan \theta =\frac{\tan {\theta }_1-\tan {\theta }_2}{1+\tan {\theta }_1\tan {\theta }_2}$.........(2)

Substituting $\tan {\theta }_1=m_1$ and $\tan {\theta }_2=m_2$ in result (2)

  $\tan \theta =\frac{\tan {\theta }_1-\tan {\theta }_2}{1+\tan {\theta }_1\tan {\theta }_2}=\frac{m_1-m_2}{1+m_1{m}_2}$

This is the formula for determining angle between two straight lines.

The slope of lines $y=x$ and $y=\frac{1}{\sqrt{3}}x$ are $m_1=1$ and $m_2=\frac{1}{\sqrt{3}}$ respectively.

Therefore, the angles between lines $y=x$ , $y=\frac{1}{\sqrt{3}}x$ ; $\theta$ is given by

  $\theta ={\tan }^{-1}\left(\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|\right)={\tan }^{-1}\left(\left|\frac{1-\frac{1}{\sqrt{3}}}{1+1\times \frac{1}{\sqrt{3}}}\right|\right)$

  $\Rightarrow \theta ={\tan }^{-1}\left(\left|\frac{\sqrt{3}-1}{\sqrt{3}+1}\right|\right)={\tan }^{-1}\left(0.268\right)$

  $\Rightarrow \theta ={15}^0$

Conclusion:

Therefore, the formula for determining angle between two straight lines is $\tan \theta =\frac{m_1-m_2}{1+m_1{m}_2}$ .The angle between lines $y=x$ and $y=\sqrt{3}x$ is ${15}^0$ .