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In right triangle ABC with \(\angle B = 90^\circ\), we have \(\sin A = 2\cos A\). What is \(\tan A\)?

 Jun 28, 2019
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sin A  =  2 cos A

                               Divide both sides of the equation by  cos A

\(\frac{\sin A}{\cos A}\)  =  2

                     And  \(\frac{\sin A}{\cos A}=\tan A\)  so we can substitute  tan A  in for  \(\frac{\sin A}{\cos A}\)

tan A  =  2

 Jun 28, 2019

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