# Question #70b87

May 13, 2017

If Quadrant I: $\frac{4}{5}$
If Quadrant II: $- \frac{4}{5}$

#### Explanation:

Use Pythagorean's Identity: ${\sin}^{2} x + {\cos}^{2} x = 1$

By substitution, we get:
${\left(\frac{3}{5}\right)}^{2} + {\cos}^{2} a = 1$

Subtract ${\left(\frac{3}{5}\right)}^{2}$ from both sides:
${\cos}^{2} a = 1 - {\left(\frac{3}{5}\right)}^{2}$
${\cos}^{2} a = 1 - \frac{9}{25}$
${\cos}^{2} a = \frac{16}{25}$

Now, we take the square root of both sides:
$\cos a = \pm \sqrt{\frac{16}{25}}$
$\cos a = \pm \frac{4}{5}$

Therefore, if $\angle A$ is in Quadrant I, $\cos a = \frac{4}{5}$
if $\angle A$ is in Quadrant II, $\cos a = - \frac{4}{5}$