# How do you find cos if sin= 5/13?

Nov 20, 2015

$\cos = \pm \frac{12}{13}$

#### Explanation:

If $\sin = \frac{5}{13}$
then
$\textcolor{w h i t e}{\text{XXX}}$ the ratio of $\left(\text{opposite side")/("hypotenuse}\right) = \frac{5}{13}$

By Pythagorean Theorem
If $\text{opposite side" = 5 " units}$ and $\text{hypotenuse" = 13 " units}$
$\textcolor{w h i t e}{\text{XXX}}$(for any $\text{units}$)
then $\text{adjacent side" = 12 " units}$
and
$\cos = \left(\text{adjacent side")/("hypotenuse}\right) = \frac{12}{13}$

However, we need to note that if the angle is in Quadrant II then the $\text{adjacent side}$ will actually be a negative value,
so
$\textcolor{w h i t e}{\text{XXX}} \cos = \frac{12}{13}$ for an angle in Q I
or
$\textcolor{w h i t e}{\text{XXX}} \cos = - \frac{12}{13}$ for an angle in Q II
(the angle can't be in Q III or Q IV, since $\sin > 0$)

May 21, 2017

$\cos x = \pm \frac{12}{13}$

#### Explanation:

Another way.
${\cos}^{2} x = 1 - {\sin}^{2} x = 1 - \frac{25}{169} = \frac{144}{169}$
$\cos x = \pm \frac{12}{13}$
sin x > 0 --> x is either in Quadrant 1 or Quadrant 2.

If x is in Quadrant 1, then, cos x > 0 --> $x = \frac{12}{13}$
If x is in Quadrant 2, then, cos < 0 --> $\cos x = - \frac{12}{13}$