ANSWERS
x=0.5 in Quadrant I.
θ = 60 degrees
x=-0.8 in Quadrant II.
θ = 143.13 degrees
x=-0.1 in Quadrant III.
θ = 264.26 degrees
x=0.8 in Quadrant IV.
θ = 323.13 degrees
Step-by-step explanation
For all the angles we know the length of the adjacent side of the triangle and the length of the hypotenuse - which is the radius of the circle.
For the first triangle in red, the angle is:
![\begin{gathered} \cos \theta=(0.5)/(1) \\ \cos \theta=0.5 \\ \theta=\cos ^(-1)0.5 \\ \theta=60º \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/bzxfxcypd27rqny8r51p43pfy9omduj283.png)
For the second triangle, in green, we can find the supplementary angle of θ:
![\begin{gathered} \cos (180º-\theta)=(0.8)/(1) \\ 180º-\theta=\cos ^(-1)0.8 \\ 180º-\theta=36.87º \\ \theta=180º-36.87º \\ \theta=143.13º \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/lrorkyysw87wigtwev6priubsg1sw636cj.png)
For the third triangle, in light-blue, the angle we'll find is (θ - 180º):
![\begin{gathered} \cos (\theta-180º)=(0.1)/(1) \\ \theta-180º=\cos ^(-1)0.1 \\ \theta=180º+84.26º \\ \theta=264.26º \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/s78nf4luy716s8rbdtn3kh2gvsknjrmqr1.png)
And for the last triangle, in pink, the angle we'll find from the triangle is (360º-θ):
![\begin{gathered} \cos (360º-\theta)=(0.8)/(1) \\ 360º-\theta=\cos ^(-1)0.8 \\ \theta=360º-36.87º \\ \theta=323.13º \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/tntemv3twxhcpephqa5sn47j7a1lpqikr1.png)