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Given [tex]sin (\theta) = \frac{1}{5}[/tex], and[tex]tan \theta = \frac{\sqrt{6} }{12}[/tex], then:
[tex]cos\theta=\frac{2\sqrt{6} }{5}[/tex]
From the details provided:
[tex]sin (\theta) = \frac{1}{5}\ \\\\tan \theta = \frac{\sqrt{6} }{12} \times \frac{\sqrt{6} }{\sqrt{6} }\\\\tan \theta = \frac{\sqrt{6} \times \sqrt{6} }{12 \times \sqrt{6} }\\\\tan \theta =\frac{1}{2\sqrt{6} }[/tex]
From the relationship above:
Opposite = 1
Adjacent = [tex]2\sqrt{6}[/tex]
Hypotenuse = 5
[tex]cos\theta=\frac{Adjacent}{Hypotenuse} \\\\cos\theta=\frac{2\sqrt{6} }{5}[/tex]
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