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

 Jun 28, 2019
 #1
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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 \(\frac{3b}{2}\).

 

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

 

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

 

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

.
 Jun 28, 2019

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