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

[tex]\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}[/tex]

[tex]\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}[/tex]
abidemiokin

The value of cos θ from the given expression is 3/5

Given the following parameters;

sin θ = 4/5

According to SOH CAH TOA

sin θ = opposite/hypotenuse

Get the hypotenuse

[tex]hyp^2 =opp^2+adj^2\\5^2 = 4^2 = adj^2\\adj^2 = 25 - 16\\adj^2 =9\\adj=3[/tex]

Get the cosine of the angle;

cos θ = adj/hyp

cos  θ = 3/5

Hence the value of cos θ from the given expression is 3/5

Learn more on SOH CAH TOA here: https://brainly.com/question/22767551

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