![\bf \pm\sqrt{(\sqrt{5})^2-1^2}=b\implies \pm\sqrt{5-1}=b\implies \pm\sqrt{4}=b\implies \pm 2=b \\\\[-0.35em] ~\dotfill\\\\ sin(\theta )=\pm\cfrac{\stackrel{opposite}{2}}{\stackrel{hypotenuse}{\sqrt{5}}}\implies \stackrel{\textit{rationalizing the denominator}}{\pm\cfrac{2}{\sqrt{5}}\cdot \cfrac{\sqrt{5}}{\sqrt{5}}\implies \pm\cfrac{2\sqrt{5}}{(\sqrt{5})^2}\implies \pm\cfrac{2\sqrt{5}}{5}}](/tpl/images/0392/5220/486f9.png)
What is sin θ when sec θ = square root of 5? rationalize the denominator if necessary.
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Mathematics, 27.11.2019 02:31 kleathers97
What is sin θ when sec θ = square root of 5? rationalize the denominator if necessary.
Answers: 3
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