a special right triangle drawn in quadrant I intersects the unit circle at point (x,y) where y = √2/2. what is the value of x in point (x,y)? use the equation of the unit circle to determine the missing coordinate
In quadrant I, the equation of the unit circle is:
x² + y² = 1
Since y = √2 / 2, we can substitute this value into the equation:
x² + (√2 / 2)² = 1
x² + 2 / 4 = 1
x² + 1 / 2 = 1
x² = 1 - 1 / 2
x² = 1 / 2
x = ± √(1 / 2)
Since we are in quadrant I, the value of x will be positive. Therefore, x = √(1 / 2).
So, the missing coordinate (x, y) is (√(1 / 2), √2 / 2).