Chapter 9.8, Problem 39E

a.

To determine

To write an equation for tan $\left({\theta }_2-{\theta }_1\right)$ in terms of $m_1$ and $m_2$ .

Answer to Problem 39E

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

Explanation of Solution

Given:

The figure shows the acute angle of intersection, ${\theta }_2-{\theta }_1$ , of two lines with slopes $m_1$ and $m_2$ .

  

Calculation:

Using the formula as follows-

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

b.

To determine

To find the acute angle of intersection of the lines $y=x-1$ and $y=\left(\frac{1}{\sqrt{3-2}}\right)x+\frac{4-\sqrt{3}}{2-\sqrt{3}}$ .

Answer to Problem 39E

  ${60}^{\circ }$

Explanation of Solution

Given:

The figure shows the acute angle of intersection, ${\theta }_2-{\theta }_1$ , of two lines with slopes $m_1$ and $m_2$ .

  

Calculation:

The first line $y=x-1$ has slope $m_1=1$

The second line $y=\left(\frac{1}{\sqrt{3-2}}\right)x+\frac{4-\sqrt{3}}{2-\sqrt{3}}$ has slope $m_2=\frac{1}{\sqrt{3}-2}$

  $\begin{array}{l}\tan ({\theta }_2-{\theta }_1)=\frac{m_2-m_1}{1+m_2{m}_1}\\ \tan ({\theta }_2-{\theta }_1)=\frac{\frac{1}{\sqrt{3}-2}-1}{1+\frac{1}{\sqrt{3}-2}}\\ \tan ({\theta }_2-{\theta }_1)=\frac{1-\sqrt{3}+2}{\sqrt{3}-2+1}\\ \tan ({\theta }_2-{\theta }_1)=\frac{3-\sqrt{3}}{\sqrt{3}-1}\\ \tan ({\theta }_2-{\theta }_1)=\frac{\sqrt{3}\left(\sqrt{3}-1)\right)}{\sqrt{3}-1}\\ {\theta }_2-{\theta }_1={\tan }^{-1}(-\sqrt{3})\\ {\theta }_2-{\theta }_1={60}^{\circ }\end{array}$