Chapter 13.2, Problem 42PPS

a.

To determine

Draw a $\Delta EAB\&\Delta ECD$ with E at the origin.

Explanation of Solution

Given:

It is given in the question that the four points are $A(-4,0),B(-4,6),C(6,0)\&D(6,8)$ .

Graph:

  

Interpretation:

In this, put all the points $A(-4,0),B(-4,6),C(6,0)\&D(6,8)$ in the graphing calculator, it will gives the required graph in proper way and the two triangles will formed.

b.

To determine

Calculate the values of the tangent of $\angle BEA\&\angle DEC$ .

Answer to Problem 42PPS

The values of the tangent of $\angle BEA\&\angle DEC$ are $\frac{3}{2}\&\frac{4}{3}$ .

Explanation of Solution

Given:

It is given in the question that the points are $A(-4,0),B(-4,6),C(6,0)\&D(6,8)$ and the graph is given below:

  

Concept Used:

In this, use the concept of trigonometric ratio i.e. $\tan \theta =\frac{p}{h}$ which is called the slope of the line or angle.

Calculation: Since, the points are $A(-4,0),B(-4,6),C(6,0)\&D(6,8)$

Now,

  $\begin{array}{l}\tan \angle BEA=\frac{\left|AB\right|}{\left|AE\right|}\\ \tan \angle BEA=\frac{6}{4}\\ \tan \angle BEA=\frac{3}{2}\end{array}$

Again,

  $\begin{array}{l}\tan \angle DEC=\frac{\left|DC\right|}{\left|CE\right|}\\ \tan \angle DEC=\frac{8}{6}\\ \tan \angle DEC=\frac{4}{3}\end{array}$

Conclusion:

The two values are $\frac{3}{2}\&\frac{4}{3}$ .

c.

To determine

Calculate the slope of BE and ED.

Answer to Problem 42PPS

The SLOPE of BE and ED are $-\frac{3}{2}\&\frac{4}{3}$ .

Explanation of Solution

Given: It is given in the question that the points are $A(-4,0),B(-4,6),C(6,0)\&D(6,8)$ and the graph is given below:

  

Concept Used:

In this, use the concept of another formula of slope i.e. $slope=\frac{Rise}{Run}$ .

Calculation: Here, the points will be $A(-4,0),B(-4,6),C(6,0)\&D(6,8)$ .

Now, by seeing the graph, Slope of BE = $\frac{Rise}{Run}=\frac{-3}{2}$

Slope of ED = $\frac{Rise}{Run}=\frac{4}{3}$

Conclusion:

The two values for the slopes are $-\frac{3}{2}\&\frac{4}{3}$ .

d.

To determine

Explain the relationship between the slope and tangent.

Answer to Problem 42PPS

The value of tangent is the same as the absolute value of the slope.

Explanation of Solution

Given:

It is given in the question that there is a tangent and a slope.

Since, the tangent gives the direction of the curve at a particular point whereas the slope also gives the direction of the point by using the positive and negative sign and with some absolute values.