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In right triangle $ABC$ with $\angle B = 90^\circ$, we have $2\sin A = 3\cos A$. What is $\tan A$?

 Apr 26, 2021
 #1
avatar+30 
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From SOHCAHTOA, we conclude that sin A = BC/AC and cos A = AB/AC. We have BC/AC=2AB/AC, or BC=2AB. You want tan A, which is BC/AB by SOHCAHTOA. But you have BC=2AB, so BC/AB=2, which is your answer.

 Apr 26, 2021
 #2
avatar+420 
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One of the trig identities state that \(\tan A = \frac{\sin A}{\cos A}\) (you could just prove it by using sohcahtoa), so that's what we need to solve for. To do that, divide both sides by cos A and then by 2 to get:

\(\frac{\sin A}{\cos A} = \boxed{\frac{3}{2}}\), which is our final answer.

textot  Apr 26, 2021

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