Chapter 6.2, Problem 6AYU

To determine

To find: y

Answer to Problem 6AYU

Solution: $$

  $y=-\frac{1}{2}$

Explanation of Solution

Given:

If (x, y) is a point on the unit circle in quadrant IV and $x=\frac{\sqrt{3}}{2}$ .

Calculation:

Unit circle is $x^2+y^2=1$ .

  ${\left(\frac{\sqrt{3}}{2}\right)}^2+y^2=1$

  $\frac{3}{4} +y^2=1$

  $y^2=1-\frac{3}{4}$

  $y^2=\frac{1}{4}$

  $y=\pm \frac{1}{2}$

The equation in the IV quadrant, y value must be negative.

Hence $y=-\frac{1}{2}$ .