Chapter 12.1, Problem 59WE

To determine

Find a first-quadrant angle $\theta ,0°<\theta <90°$ ,for which an angle six times as large as $\theta$ will be in the given quadrant.

Answer to Problem 59WE

  $\theta =40°$

Explanation of Solution

Given:

Third quadrant

Calculation:

Given $6\theta$ is in third quadrant.

So,

  $\begin{array}{l}180°<6\theta <270°\\ \Rightarrow \frac{180°}{6}<\frac{6\theta }{6}<\frac{270°}{6}\mathrm{...............}[\text{divide by 6]}\\ \Rightarrow 30°<\theta <45°\end{array}$

So, let $\theta =40°$

Check $6\theta =240°$ is in third quadrant.

So, the solution is $\theta =40°$ .