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# trig

+1
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+500

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
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0

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
+327
+2

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