The last of my points. Given the unit circle, what is the value of y? This my last question.

Answer:
y = - [tex]\frac{\sqrt{111} }{20}[/tex]
Step-by-step explanation:
using Pythagoras' identity in the right triangle formed by the radius 1 and the given point, then
y² + (- [tex]\frac{17}{20}[/tex] )² = 1²
y² + [tex]\frac{289}{400}[/tex] = 1
y² = 1 - [tex]\frac{289}{400}[/tex] = [tex]\frac{111}{400}[/tex] ( take square root of both sides )
y = ± [tex]\sqrt{\frac{111}{400} }[/tex] = ± [tex]\frac{\sqrt{111} }{20}[/tex]
since the point is in the third quadrant then y < 0
y = - [tex]\frac{\sqrt{111} }{20}[/tex]