home > s25 ma2090 - dambroise precalculus > assessment\nhomework #12 - circular sine and cosine\nscore…

Explanation:

Step1: Recall unit - circle equation

The equation of the unit circle is $x^{2}+y^{2}=1$.

Step2: Substitute given $y$ - value

We know $y =-\frac{3}{5}$, so $x^{2}+\left(-\frac{3}{5}\right)^{2}=1$.

Step3: Simplify the equation

$x^{2}+\frac{9}{25}=1$, then $x^{2}=1 - \frac{9}{25}=\frac{25 - 9}{25}=\frac{16}{25}$.

Step4: Solve for $x$

$x=\pm\sqrt{\frac{16}{25}}=\pm\frac{4}{5}$.

Step5: Determine the sign of $x$ based on the quadrant

Since the point $P$ is in quadrant IV where $x>0$, so $x = \frac{4}{5}$.

Answer:

$\frac{4}{5}$