Given that the angle $ heta$ in standard position intersects the unit circle at the point $(x, y)$ in the first quadrant where $x = frac{3}{5}$, find the $y$-coordinate of the point. Use the Pythagorean identity for the unit circle to show your work.

Given the Pythagorean identity for the unit circle:

$ x^2 + y^2 = 1 $

where $ x = \frac{3}{5}$, substitute this value into the identity:

$ \left( \frac{3}{5} \right)^2 + y^2 = 1 $

$ \frac{9}{25} + y^2 = 1 $

Subtract $ \frac{9}{25}$ from both sides:

$ y^2 = 1 – \frac{9}{25} $

$ y^2 = \frac{25}{25} – \frac{9}{25} $

$ y^2 = \frac{16}{25} $

Taking the square root of both sides:

$ y = \pm \sqrt{\frac{16}{25}} $

$ y = \pm \frac{4}{5} $

Since (x, y) is in the first quadrant:

$ y = \frac{4}{5} $

Given that $x = frac{3}{5}$ and using the unit circle identity $x^2 + y^2 = 1$:

$ left( frac{3}{5}
ight)^2 + y^2 = 1 $

$ frac{9}{25} + y^2 = 1 $

Subtract $ frac{9}{25}$ from both sides:

$ y^2 = 1 – frac{9}{25} $

$ y^2 = frac{16}{25} $

Take the square root of both sides:

$ y = pm frac{4}{5} $

Since the point is in the first quadrant, $ y > 0 $:

$ y = frac{4}{5} $

Using $x = frac{3}{5}$ and the unit circle identity $x^2 + y^2 = 1$:

$ left( frac{3}{5}
ight)^2 + y^2 = 1 $

$ frac{9}{25} + y^2 = 1 $

$ y^2 = frac{16}{25} $

$ y = frac{4}{5} $

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