Suppose that θ is an acute angle of a right triangle and that sec(θ) = (5)/(2). Find cos(θ) and

Okay, we're given that the cosine of an angle is the square root of 21 over 7, and we're being a

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Suppose that \(\theta\) is an acute angle of a right triangle and that \(\sec(\theta) = \frac{5}{2}\). Find \(\cos(\theta)\) and \(\csc(\theta)\). \(\circ \cos(\theta) = \frac{5\sqrt{21}}{21}, \csc(\theta) = \frac{\sqrt{21}}{2}\) \(\circ \cos(\theta) = \frac{2}{5}, \csc(\theta) = \frac{\sqrt{21}}{5}\) \(\circ \cos(\theta) = \frac{2}{5}, \csc(\theta) = \frac{5\sqrt{21}}{21}\) \(\circ \cos(\theta) = \frac{\sqrt{21}}{5}, \csc(\theta) = \frac{2\sqrt{21}}{21}\) \(\circ \cos(\theta) = \frac{5}{2}, \csc(\theta) = \frac{5\sqrt{21}}{21}\)

          Suppose that \(\theta\) is an acute angle of a right triangle and that \(\sec(\theta) = \frac{5}{2}\). Find \(\cos(\theta)\) and \(\csc(\theta)\).

\(\circ \cos(\theta) = \frac{5\sqrt{21}}{21}, \csc(\theta) = \frac{\sqrt{21}}{2}\)

\(\circ \cos(\theta) = \frac{2}{5}, \csc(\theta) = \frac{\sqrt{21}}{5}\)

\(\circ \cos(\theta) = \frac{2}{5}, \csc(\theta) = \frac{5\sqrt{21}}{21}\)

\(\circ \cos(\theta) = \frac{\sqrt{21}}{5}, \csc(\theta) = \frac{2\sqrt{21}}{21}\)

\(\circ \cos(\theta) = \frac{5}{2}, \csc(\theta) = \frac{5\sqrt{21}}{21}\)

Suppose that θ is an acute angle of a right triangle and that sec(θ) = (5)/(2). Find cos(θ) and csc(θ).

∘cos(θ) = (5√(21))/(21), csc(θ) = (√(21))/(2)

∘cos(θ) = (2)/(5), csc(θ) = (√(21))/(5)

∘cos(θ) = (2)/(5), csc(θ) = (5√(21))/(21)

∘cos(θ) = (√(21))/(5), csc(θ) = (2√(21))/(21)

∘cos(θ) = (5)/(2), csc(θ) = (5√(21))/(21)

Added by Tara C.

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