Chapter 13.6, Problem 29PPS

a.

To determine

To find: Draw lines representing ${30}^0,{60}^0,{150}^0,{210}^0,{315}^0$ .

Explanation of Solution

  

This is for $\theta ={30}^0$

  

This is for $\theta ={60}^0$

  

This is for $\theta ={150}^0$

  

This is for $\theta ={210}^0$

  

This is for $\theta ={315}^0$

b.

To determine

To find: Use a table to values to show the slope of each line to the nearest tenth.

Explanation of Solution

Angle	Slope
${30}^0$	$0.6$
${60}^0$	$1.7$
${120}^0$	$-1.7$
${150}^0$	$-0.6$
${210}^0$	$0.6$
${315}^0$	$-1$

c.

To determine

To find: What conclusions can we make about the relationship between the terminal side of the angle and the slope.

Answer to Problem 29PPS

Slope $=\tan \left(\theta \right)$

Explanation of Solution

The slope corresponds to the tangent of the angle.

For

  $\theta ={120}^0$ ,

The x-coordinate of P is $\frac{-1}{2}$ and y-coordinate is $\frac{\sqrt{3}}{2}$ .

Slope $=\frac{y}{x}$

On substituting values

We get

  $\Rightarrow$ Slope $=\frac{\left(\frac{\sqrt{3}}{2}\right)}{\left(\frac{-1}{2}\right)}$

  $\Rightarrow$ Slope $=\left(\frac{\sqrt{3}}{2}\right)\times \left(-2\right)$

  $\Rightarrow$ Slope $=-\sqrt{3}$

This means

Slope $=\tan \left(\theta \right)$