Answer:
[tex]\cos(\theta) = -\frac{\sqrt{17}}{6}[/tex]
Step-by-step explanation:
Given
[tex]\tan(\theta) = -\sqrt{\frac{19}{17}}[/tex]
Required
Determine [tex]\cos(\theta)[/tex]
We have:
[tex]\tan(\theta) = -\sqrt{\frac{19}{17}}[/tex]
Split
[tex]\tan(\theta) = -\frac{\sqrt{19}}{\sqrt{17}}[/tex]
tan is calculated as:
[tex]\tan(theta) = \frac{opposite}{adjacent}[/tex]
So:
[tex]Opposite = -\sqrt{19[/tex]
[tex]Adjacent = \sqrt{17[/tex]
And:
[tex]Hypotenuse^2 = Opposite^2 + Adjacent^2[/tex] --- Pythagoras theorem
[tex]Hypotenuse^2 = (-\sqrt{19})^2 + (\sqrt{17})^2[/tex]
[tex]Hypotenuse^2 = 19 + 17[/tex]
[tex]Hypotenuse^2 = 36[/tex]
Take square roots
[tex]Hypotenuse = 6[/tex]
[tex]\cos(\theta) = \frac{Adjacent}{Hypotenuse}[/tex]
[tex]\cos(\theta) = \frac{\sqrt{17}}{6}[/tex]
Since it is in the second quadrant, then:
[tex]\cos(\theta) = -\frac{\sqrt{17}}{6}[/tex]
