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Find the exact value of sec()θfor the angle in standard position passing through the point (-3, 2) on its terminal side.

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well, we know the angle θ terminal location, is at -3,2, so the adjacent side is -3 and the opposite side is 2, so it looks like the picture below, therefore


\bf (\stackrel{a}{-3}~,~\stackrel{b}{2})\impliedby \textit{let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=√(a^2+b^2) \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=√((-3)^2+2^2)\implies c=√(13)\qquad \qquad \qquad sec(\theta )=\cfrac{\stackrel{hypotenuse}{√(13)}}{\stackrel{adjacent}{-3}}
Find the exact value of sec()θfor the angle in standard position passing through the-example-1
User Tommizzle
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