![\bf cos(\theta )=\cfrac{\stackrel{adjacent}{-5}}{\stackrel{hypotenuse}{8}}\qquad \impliedby \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{8^2-(-5)^2}=b\implies \pm\sqrt{64-25}=b\implies \pm\sqrt{39}=b\implies \stackrel{II~Quadrant}{+\sqrt{39}=b} \\\\[-0.35em] ~\dotfill](/tpl/images/0186/4228/c935c.png)
To be an angle in quadrant ii such that cos o= -5/8
find the exact values of csc o and cot o....
Mathematics, 22.08.2019 01:20 MARTINASLACK299
To be an angle in quadrant ii such that cos o= -5/8
find the exact values of csc o and cot o.
Answers: 1
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