Suppose that is a point in quadrant lying on the unit circle. Find . Write the exact value, not a decimal

Suppose that (3/5,y) is a point in quadrant IV lying on the unit circle.

Find y. Write the exact value, not a decimal approximation.

Note that the equation of the unit circle is:
x^2 + y^2 = 1.

Since (3/5, y) lies on the circle, we can plug it in to determine that:
(3/5)^2 + y^2 = 1
==> y^2 = 1 - (3/5)^2 = 16/25
==> y = ±4/5.

In Quadrant IV, y < 0, so y = -4/5.

I hope this helps!

recall that explicitly solving for y in x^2+y^2=1 yields y=+(1-x^2)^(1/2) and y=-(1-x^2)^(1/2). since quadrant iv is part of the lower side, or second function of the unit circle, we just plug in (3/5) into x. thus, we obtain
y=(1-(3/5)^2)^(1/2)
y=(1-(9/25))^(1/2)
y=(25/25-9/25)^(1/2)
y=(16)^(1/2)
y=(+/-)4 BUT remember that this answer must only satisfy answers in quadrant iv. thus, the answer is (3/5,-4), since quadrant iv contains negative values of y .

if you need any more help you can contact me
-jake

well if the cos is 3/5 in quadrant 4,sin will be negative,the hyptoenuse is 1 always is in the Unit Circle

(3/5)^2+b^2=1
9/25 +b^2 =1
25/25-9/25=b^2
16/25=b^2
b=+-4/5
b=-4/5