+1206 In right triangle \(ABC\) with \(\angle B = 90^\circ\), we have \(2\sin A = 3\cos A\). What is \(\tan A\)?
+997 STEP 1.
So, we have a right triangle \(\triangle {ABC}\) with \(\angle B = 90^\circ\).
Let sides \({AC} = a\), \(AB = b\). and \(CB = c\).
\(2\sin A\) is equal to \(2(\frac{c}{a})\), and \(3\cos A \) is equal to \(3(\frac{b}{a})\)
Therfore,
\(2(\frac{c}{a}) = 3(\frac{b}{a})\),
\(\frac{2c}{a} = \frac{3b}{a}\),
\(2c = 3b\).
STEP 2.
\(\tan A\) is equal to \(\frac{c}{b}\).
We can substitute \(c\) for \(\frac{3b}{2}\).
\(\tan A\) \(=\) \(\frac{\frac{3b}{2}}{b}\),
\(\tan A\) \(=\) \(\frac{3b}{2} \times \frac{1}{b}\),
\(\tan A\) \(=\) \(\boxed{\frac{3}{2}}\)
