Question

If sin θ = 4/5 and cos θ is in quadrant II, then cos θ =

Answer 1

Answer:

It’s -3/5 :)

Step-by-step explanation:

Answer 2

Keeping in mind that in the II quadrant, since is positive and the cosine is negative, then,

$\bf sin(\theta )=\cfrac{\stackrel{opposite}{4}}{\stackrel{hypotenuse}{5}}\impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a\qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}$

$\bf \pm\sqrt{5^2-4^2}=a\implies \pm\sqrt{25-16}=a\implies \pm\sqrt{9}=a \\\\\\ \pm 3=a\implies \stackrel{\textit{II~quadrant}}{-3} \\\\\\ cos(\theta )=\cfrac{adjacent}{hypotenuse}\qquad \qquad \qquad cos(\theta )=\cfrac{-3}{5}$