A special right triangle drawn in Quadrant 1 intersects the unit circle at point (x,y) where y = sqrt 2/2. What is the value of x

Answer 1

To find the value of $x$ for the point $(x,y)$ on the unit circle where $y=\frac{\sqrt{2}}{2}$, we start with the equation of the unit circle:

$$x^2+y^2=1$$

Given that $y=\frac{\sqrt{2}}{2}$, we substitute $y$ into the equation of the circle:

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

Calculating ${\left(\frac{\sqrt{2}}{2}\right)}^2$:

$${\left(\frac{\sqrt{2}}{2}\right)}^2=\frac{2}{4}=\frac{1}{2}$$

Now we can substitute this back into the circle's equation:

$$x^2+\frac{1}{2}=1$$

To isolate $x^2$, subtract $\frac{1}{2}$ from both sides:

$$x^2=1-\frac{1}{2}=\frac{1}{2}$$

Now we take the square root of both sides to find $x$:

$$x=\sqrt{\frac{1}{2}}\text{or}x=-\sqrt{\frac{1}{2}}$$

Since we are in Quadrant 1, where both $x$ and $y$ are positive, we have:

$$x=\sqrt{\frac{1}{2}}=\frac{\sqrt{2}}{2}$$

So the value of $x$ is:

$$\boxed{{\displaystyle \frac{\sqrt{2}}{2}}}$$