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Question

The point  is on the unit circle. If the y-coordinate of  is , and  is in quadrant II , then

Answer 1

P = (x,y) = (x,23) on the unit circle where hypotenuse = 1 = sqr(x^2 +y^2) = sqr(x^2 +23^2)

1 = x^2 + 23^2

x^2 =1- 529=-528

x = +/-isqr528 = about -/+22.98i

either you mis-copied the problem, the problem is misprinted or

you're in an imaginary unit circle

if the problem had been y = .23, then

x = +sqr(1-.23^2) = sqr(1-.0529)= sqr.9471 = about .9732

P = about (.9732, .23)