![\bf \pm\sqrt{5^2-(-1)^2}=a\implies \pm\sqrt{24}=a\implies \stackrel{III~Quadrant}{-\sqrt{24}=a}\implies -2\sqrt{6}=a \\\\[-0.35em] ~\dotfill\\\\ ~\hfill cos(\theta )=\cfrac{\stackrel{adjacent}{-2\sqrt{6}}}{\stackrel{hypotenuse}{5}}~\hfill](/tpl/images/0160/1263/ef0fa.png)
Hey i'm not too sure about this. what would the exact trig value be? .
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Mathematics, 02.08.2019 02:20, med69
Hey i'm not too sure about this. what would the exact trig value be? .
Answers: 1
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