Question

a. Show that tan∠A = slope of AC.

b. Use trigonometry to find m∠A.

Short answer

1. We showed that$\tan \angle A=\text{slopeof}\overline{AC}$ .
2. the value of$\angle A={38.65}^{\circ }$ .

Step-by-step solution

Step-1 – Given

Given figure is:

Step-2 – To determine

We have to show that $\tan \angle A=\text{slopeof}\overline{AC}$.

Step-3 – Calculation 

Slope formula with two points$(x_1,y_1)\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}(x_2,y_2)$ is:

$m=\frac{y_2-y_1}{x_2-x_1}$

The given points are$A(0,0)$ and$C(5,4)$ .

$m=\frac{4-0}{5-0}=\frac{4}{5}=0.8$

Also,

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

Therefore, we proved that$\tan \angle A=\text{slopeof}\overline{AC}$ .

Step-1 – Given

Given figure is:

Step-2 – To determine

We have to find$m\angle A$ .

Step-3 – Calculation 

Since,

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

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