# Given a unit circle, what is the value of y in the first quadrant corresponding to an x-coordinate of 5/8?

Aug 2, 2018

$\frac{\sqrt{39}}{8}$ or $\approx 0.78$

#### Explanation:

In a unit circle, the $x$ coordinate represents the $\cos$ value, and the $y$ coordinate represents the $\sin$ value. Thus, we can say

$\cos x = \frac{5}{8}$

From SOH-CAH-TOA, we know that this means if we have a right triangle, the adjacent side is $5$ and the hypotenuse is $8$.

We can find the $\sin$ value using the Pythagorean Theorem

${a}^{2} + {b}^{2} = {c}^{2}$

We can plug our values in to get

${5}^{2} + {b}^{2} = {8}^{2}$

$\implies {b}^{2} + 25 = 64$

$\implies {b}^{2} = 39 \implies b = \sqrt{39}$

We know that sine is equal to opposite over the hypotenuse. We essentially just found the opposite side, so we can plug in to get

$\sin x = \frac{\sqrt{39}}{8}$, which is approximately $0.78$.

This is the $y$ coordinate.

Hope this helps!

Aug 3, 2018

$y = 0.78$

#### Explanation:

Picture a right triangle with corners made up of the point on the circle (whose x coordinate is 5/8) and the projections of that point on the x and y axes. The point on the x axis is 5/8 of the way from the origin to the circle. Since this is a unit circle, the hypotenuse by definition = 1.

Using Pythagoras:
${x}^{2} + {y}^{2} = {h}^{2}$
where x and y are the coordinates of the spot on the unit circle and h is the length of the hypotenuse which, as I said, is 1.

${\left(\frac{5}{8}\right)}^{2} + {y}^{2} = {1}^{2}$

${0.625}^{2} + {y}^{2} = 0.391 + {y}^{2} = 1$

$y = \sqrt{1 - .391} = 0.78$

Another approach:
Let the angle between a line from the origin to the spot on the unit circle be $\theta$. The x and y coordinates are

$x = 1 \cdot \cos \theta \mathmr{and} y = 1 \cdot \sin \theta$.

It is given that x = 5/8.

$\theta = {\cos}^{-} 1 \left(\frac{5}{8}\right) = {51.3}^{\circ}$

$y = 1 \cdot \sin \theta = 1 \cdot \sin {51.3}^{\circ} = 0.78$

I hope this helps,
Steve