Chapter 13.2, Problem 30WE

a.

To determine

To show that $\tan \angle A=\text{slope of }\overline{AC}$

Answer to Problem 30WE

It is proved that $\tan \angle A=\text{slope of }\overline{AC}$

Explanation of Solution

Given:

Below is the figure given

  

Calculation:

Slope of a line when two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are given is

  $m=\frac{y_2-y_1}{x_2-x_1}$

Points given are $A\left(0,0\right)$ and $C\left(5,4\right)$

  $m=\frac{4-0}{5-0}=\frac{4}{5}=0.8$

Also,

  $\tan \angle A=\frac{\text{perpendicular}}{\text{Base}}=\frac{4}{5}=0.8$

Hence, it is proved that $\tan \angle A=\text{slope of }\overline{AC}$ .

b.

To determine

To find $m\angle A$

Answer to Problem 30WE

Explanation of Solution

Given:

Below is the figure given

  

Calculation:

Since,

  $\begin{array}{l}\tan \angle A=\frac{\text{perpendicular}}{\text{Base}}=\frac{4}{5}=0.8\\ \tan \angle A=0.8\\ \angle A={\tan }^{-1}\left(0.8\right)\\ \angle A={38.65}^{\circ }\end{array}$

Therefore , the value of $\angle A={38.65}^{\circ }$ .