In right triangle ABC with \(\angle B = 90^\circ\), we have \(\sin A = 2\cos A\). What is \(\tan A\)?
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
+1206 
