Chapter 7.5, Problem 107AYU

To determine

To find: $\tan \theta =\frac{m_2-m_1}{1+m_2\cdot m_1}$

Answer to Problem 107AYU

Solution:

$\tan \theta =\frac{m_2-m_1}{1+m_2\cdot m_1}$ is shown.

Explanation of Solution

Given:

$L_1\text{ and }{L}_2$ are two non-vertical intersecting lines.

$\Theta =$ acute angle between $L_1\text{ and }{L}_2$.

$m_1\text{ and }{m}_2$ are slopes of $L_1\text{ and }{L}_2$.

Calculation:

$m_1=\tan {\theta }_1$

$m_2=\tan {\theta }_2$

External angle property,

${\theta }_2-{\theta }_1=\theta$

$\text{tan}({\theta }_2-{\theta }_1)=\text{ tan}\theta$

By difference formula of $\text{tan}$,

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

$\text{LHS }=\text{ RHS}$