Respuesta :
∅ = 60
∅ = 45
∅ = 60
∅ = 30
For a 300° angle, it lies in Quadrant IV; therefore we subtract it from 360:
360-300 = 60
For a 225° angle, it lies in Quadrant III; therefore we subtract 180 from it:
225-180 = 45
For a 480° angle, it will lie in Quadrant II (it will lie in the same place as 120°); therefore we subtract it from 180:
180-120 = 60
For a -210° angle, it will lie in Quadrant II (it will lie in the same place as 150°); therefore we subtract it from 180:
180-150 = 30
∅ = 45
∅ = 60
∅ = 30
For a 300° angle, it lies in Quadrant IV; therefore we subtract it from 360:
360-300 = 60
For a 225° angle, it lies in Quadrant III; therefore we subtract 180 from it:
225-180 = 45
For a 480° angle, it will lie in Quadrant II (it will lie in the same place as 120°); therefore we subtract it from 180:
180-120 = 60
For a -210° angle, it will lie in Quadrant II (it will lie in the same place as 150°); therefore we subtract it from 180:
180-150 = 30
Let [tex]\phi[/tex] denote the reference angle, and [tex]\theta[/tex] the given angle, then
[tex]\text{When }\theta=300^{\circ},\phi=60^{\circ}\\\text{When }\theta=225^{\circ},\phi=45^{\circ}\\\text{When }\theta=480^{\circ},\phi=60^{\circ}\\\text{When }\theta=-210^{\circ},\phi=30^{\circ}[/tex]
The reference angle for a given angle equals the smallest positive angle that can be formed by the terminal side of the given angle and the x-axis.
When [tex]\theta=300^{\circ}[/tex], it lies in the fourth quadrant,
For the fourth quadrant, we have
[tex]\phi=(360-\theta)^{\circ}\\=(360-300)^{\circ}\\=60^{\circ}[/tex]
When [tex]\theta=225^{\circ}[/tex], it lies in the third quadrant,
For the third quadrant, we have
[tex]\phi=(\theta-180)^{\circ}\\=(225-180)^{\circ}\\=45^{\circ}[/tex]
When [tex]\theta=480^{\circ}[/tex], it lies in the second quadrant,
(since the remainder after dividing [tex]480 \text{ by }360[/tex] is [tex]120^{\circ}[/tex], and second quadrant)
For the second quadrant, we have
[tex]\phi=(180-\theta)^{\circ}\\=(180-120)^{\circ}\\=60^{\circ}[/tex]
When [tex]\theta=-210^{\circ}[/tex], it lies in the second quadrant,
(since the equivalent positive angle of [tex]-210 \text{ is }360+(-210)=150[/tex] , and [tex]150^{\circ}[/tex] lies in the second quadrant)
For the second quadrant, we have
[tex]\phi=(180-\theta)^{\circ}\\=(180-150)^{\circ}\\=30^{\circ}[/tex]
Learn more about reference angles here: https://brainly.com/question/7099725
