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nobillionaireNobley
Given a right triangle with hypothenus of measure 34, the side opposite the angle θ of measure 30, and the side adjacent the angle theta of measure 16.

[tex]\sin\theta= \frac{opposite}{hypothenuse} = \frac{30}{34} = \frac{15}{17} \\ \\ \Rightarrow \theta=\sin^{-1}\left( \frac{15}{17} \right) \\ \\ \\ \cos\theta= \frac{adjacent}{hypothenuse} = \frac{16}{34} = \frac{8}{17} \\ \\ \Rightarrow \theta=\cos^{-1}\left( \frac{8}{17} \right) \\ \\ \\ \tan\theta= \frac{opposite}{adjacent} = \frac{30}{16} = \frac{15}{8} \\ \\ \Rightarrow \theta=\cos^{-1}\left( \frac{15}{8} \right)[/tex]

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