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Find an angle e in the interval [0°, 90°) that satisfies each statement. Give answer to the nearest TENTH of a degree. a) sin= 0.3697 degrees b) cos= 0.7265 degrees degrees c) sec= 2.3232 degrees d) csc= 1.1234

User Emetiel
by
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1 Answer

4 votes

first at all, we are gonna solve this using inverse trigonometric functions

like this:


\begin{gathered} \sin \alpha\text{ = x} \\ \alpha\text{ = }\sin ^(-1)x \end{gathered}

for the first case:

a)


\begin{gathered} \sin \alpha\text{ = }0.3697 \\ \alpha=\sin ^(-1)(0.3697) \\ \alpha=21.697\ldots \\ \alpha\cong20 \end{gathered}

b)


\begin{gathered} \cos \alpha\text{ = }0.7265 \\ \alpha=\cos ^(-1)(0.7265) \\ \alpha=43.406\ldots \\ \alpha\cong40 \end{gathered}

c)

In this case, we are going to use the following trigonometric identity:


\begin{gathered} \sec \alpha=(1)/(\cos \alpha) \\ \end{gathered}

later,


\begin{gathered} \sec \alpha=2.3232 \\ (1)/(\cos \alpha)=2.3232 \\ \cos \alpha=(1)/(2.3232) \\ \alpha=\cos ^(-1)((1)/(2.3232)) \\ \alpha=64.504\ldots \\ \alpha\cong60 \end{gathered}

d)

In this case, we are going to use the following trigonometric identity:


\begin{gathered} csc\alpha=(1)/(sin\alpha) \\ \end{gathered}

later,


\begin{gathered} \csc \alpha=1.1234 \\ (1)/(sin\alpha)=1.1234 \\ \sin \alpha=(1)/(1.1234) \\ \alpha=\sin ^(-1)((1)/(1.1234)) \\ \alpha=62.892\ldots \\ \alpha\cong60 \end{gathered}

finally, those are your answers.

User Peter DeGlopper
by
7.9k points
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