For the rotation -152^(@), find the coterminal angle from 0^(@) <= theta < 360^(@), the quadrant, and the reference angle. The coterminal angle is ◻^(@), which lies in Quadrant ◻, with a reference angle of ◻^(@).

Add 360°: To find the coterminal angle, add 360° to -152° until the result is between 0° and 360°. -152° + 360° = 208°Check result: Check if 208° is between 0° and 360°. Yes, it is.Determine quadrant: Determine the quadrant for 208°. Since 208° is more than 180° but less than 270°, it's in Quadrant III.Find reference angle: Find the reference angle for 208° in Quadrant III. Reference angle = 180° - (208° - 180°) Reference angle = 180° - 28° Reference angle = 152°

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For the rotation
-152^(@), find the coterminal angle from
0^(@) <= theta < 360^(@), the quadrant, and the reference angle.
The coterminal angle is
◻^(@), which lies in Quadrant
◻, with a reference angle of
◻^(@).

For the rotation $-152^{\circ}$ , find the coterminal angle from 0^{\circ} \leq \theta<360^{\circ} , the quadrant, and the reference angle. $\newline$ The coterminal angle is $\square^{\circ}$ , which lies in Quadrant $\square$ , with a reference angle of $\square^{\circ}$ .

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Q. For the rotation

-152^{\circ}

, find the coterminal angle from

0^{\circ} \leq \theta<360^{\circ}

, the quadrant, and the reference angle.

\newline

The coterminal angle is

\square^{\circ}

, which lies in Quadrant

\square

, with a reference angle of

\square^{\circ}

Add $360$ °: To find the coterminal angle, add $360°$ to $-152°$ until the result is between $0°$ and $360°$ . $\newline$ $-152° + 360° = 208°$
Check result: Check if $208^\circ$ is between $0^\circ$ and $360^\circ$ . $\newline$ Yes, it is.
Determine quadrant: Determine the quadrant for $208°$ . Since $208°$ is more than $180°$ but less than $270°$ , it's in Quadrant III.
Find reference angle: Find the reference angle for $208°$ in Quadrant III. $\newline$ Reference angle = $180° - (208° - 180°)$ $\newline$ Reference angle = $180° - 28°$ $\newline$ Reference angle = $152°$