If sin A = 3/5 and A lies in the 2nd quadrant, what is the value of tan A?
-3/4
Given \(\sin A = \frac{3}{5}\), using the Pythagorean identity, \(\cos A = \pm\sqrt{1-\sin^2 A} = \pm\sqrt{1-\frac{9}{25}} = \pm\frac{4}{5}\).
Since A lies in the 2nd quadrant, cosine is negative there, so \(\cos A = -\frac{4}{5}\).
Then \(\tan A = \frac{\sin A}{\cos A} = \frac{3/5}{-4/5} = -\frac{3}{4}\).
Hence, the answer is -3/4.
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