Chapter 12.2, Problem 1AGP

To determine

To find: Draw an angle with the given measure in standard position.

Answer to Problem 1AGP

Reference angle $={\theta }^{\text{'}}={80}^0$

Explanation of Solution

Given information:

  $\theta ={80}^0$

Concept:

If $\theta$ is the non-quadrantal angle in standard position, its reference angle ${\theta }^{\text{'}}$ is the acute angle formed by the terminal side of $\theta$ and the x-axis.

The ruled for finding the measures of reference angles for $0^0<\theta <{360}^0$ or $0^0<\theta <2\pi$ are shown

For Quadrant − I

  ${\theta }^{\text{'}}=\theta$

For Quadrant − II

  ${\theta }^{\text{'}}={180}^0-\theta$ or ${\theta }^{\text{'}}=\pi -\theta$

For Quadrant − III

  ${\theta }^{\text{'}}=\theta -{180}^0$ or ${\theta }^{\text{'}}=\theta -\pi$

For Quadrant − IV

  ${\theta }^{\text{'}}={360}^0-\theta$ or ${\theta }^{\text{'}}=2\pi -\theta$

Given

  $\theta ={80}^0$

The terminal side of $\theta ={80}^0$ lies in Quadrant − I

Therefore,

  $\begin{array}{l}{\theta }^{\text{'}}=\theta \\ \Rightarrow {\theta }^{\text{'}}={80}^0\end{array}$

Draw the terminal side of the angle ${80}^0$ counterclockwise from the positive x-axis.

Therefore,

Reference angle $={\theta }^{\text{'}}={80}^0$

Sketching the given angle is: