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Marina CMI [18]
3 years ago
11

The angle θ lies in Quadrant II .

Mathematics
1 answer:
Andreyy893 years ago
6 0

let's keep in mind that, in the II Quadrant, cosine is negative and sine, is positive.

cosine is adjacent/hypotenuse, however the hypotenuse is simply a radius unit, and thus is never negative, so in the -(2/3) the negative must be the numerator, -2.


\bf cos(\theta )=\cfrac{\stackrel{adjacent}{-2}}{\stackrel{hypotenuse}{3}}\impliedby \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{3^2-(-2)^2}=b\implies \pm\sqrt{5}=b\implies \stackrel{\textit{II Quadrant}}{+\sqrt{5}=b}~\hfill tan(\theta )=\cfrac{\stackrel{opposite}{\sqrt{5}}}{\stackrel{adjacent}{-2}} \\\\\\ ~\hspace{34em}

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Answer:

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The graph shows a system consisting of a linear equation and a quadratic equation. What is the solution(s) to the system?
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