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

Logic Jun 28, 2019

#1**+7 **

**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 \(\frac{3b}{2}\).

\(\tan A\) \(=\) \(\frac{\frac{3b}{2}}{b}\),

\(\tan A\) \(=\) \(\frac{3b}{2} \times \frac{1}{b}\),

\(\tan A\) \(=\) \(\boxed{\frac{3}{2}}\)

.KnockOut Jun 28, 2019