We know that π radians is equivalent to 180°. To solve this problem, we need to transform the given measurements from radians to degrees. Therefore:
FIRST.
[tex]\frac{23\pi}{4}=? \\ \\ \pi ------>180^{\circ} \\ \\ \frac{23\pi}{4}----->x \\ \\ \\ x=\frac{23\pi}{4}\left(\frac{180^{\circ}}{\pi}\right) \\ \\ \therefore x=1035^{\circ}[/tex]
360° implies a full rotation. So, in 1035° there are:
[tex]Rotations=\frac{1035^{\circ}}{360^{\circ}}=2 \ \frac{7}{8} \ rotations[/tex]
If there are 2 full rotations here, the equivalent measurements less than or equal to 360° can be found as follows:
[tex]\alpha = 1035^{\circ}-2\times 360^{\circ} \\ \\ \therefore \boxed{\alpha =315^{\circ}}[/tex]
SECOND.
[tex]\frac{18\pi}{5}=? \\ \\ \pi ------>180^{\circ} \\ \\ \frac{18\pi}{5}----->x \\ \\ \\ x=\frac{18\pi}{5}\left(\frac{180^{\circ}}{\pi}\right) \\ \\ \therefore x=648^{\circ}[/tex]
In 648° there are:
[tex]Rotations=\frac{648^{\circ}}{360^{\circ}}=1 \ \frac{4}{5} \ rotations[/tex]
There are 1 full rotation, so the equivalent measurements less than or equal to 360° is:
[tex]\alpha = 648^{\circ}-360^{\circ} \\ \\ \therefore \boxed{\alpha=288^{\circ}}[/tex]
THIRD.
[tex]\frac{22\pi}{9}=? \\ \\ \pi ------>180^{\circ} \\ \\ \frac{22\pi}{9}----->x \\ \\ \\ x=\frac{22\pi}{9}\left(\frac{180^{\circ}}{\pi}\right) \\ \\ \therefore x=440^{\circ}[/tex]
In 440° there are:
[tex]Rotations=\frac{440^{\circ}}{360^{\circ}}=1 \ \frac{2}{9} \ rotations[/tex]
There are 1 full rotation, so the equivalent measurements less than or equal to 360° is:
[tex]\alpha = 440^{\circ}-360^{\circ} \\ \\ \therefore \boxed{\alpha=80^{\circ}}[/tex]
FOURTH.
[tex]\frac{19\pi}{3}=? \\ \\ \pi ------>180^{\circ} \\ \\ \frac{19\pi}{3}----->x \\ \\ \\ x=\frac{19\pi}{3}\left(\frac{180^{\circ}}{\pi}\right) \\ \\ \therefore x=1440^{\circ}[/tex]
In 1440° there are:
[tex]Rotations=\frac{1440^{\circ}}{360^{\circ}}=3 \ \frac{1}{6} \ rotations[/tex]
There are 3 full rotations, so the equivalent measurements less than or equal to 360° is:
[tex]\alpha = 1440^{\circ}-3\times 360^{\circ} \\ \\ \therefore \boxed{\alpha=60^{\circ}}[/tex]
Finally, the answers are:
[tex]Pairs \\ \\ \\ 60^{\circ} \rightarrow \frac{19\pi}{3} \\ \\ 288^{\circ} \rightarrow \frac{18\pi}{5} \\ \\ 315^{\circ} \rightarrow \frac{23\pi}{4} \\ \\ 80^{\circ} \rightarrow \frac{22\pi}{9}[/tex]
