5 \text { If } \cos A=\frac{4}{5} \text { and } A \text { is in Quadrant I, what is the value of } \sin A \cdot \tan A \text { ? } \( \begin{array}{llll}\text { 1) } \frac{9}{20} & \text { 2) } \frac{12}{25} & \text { 3) } \frac{16}{25} & \text { 4) } \frac{16}{20}\end{array} \)

To find \(\sin A \cdot \tan A\), we start with the given \(\cos A = \frac{4}{5}\). Since \(A\) is in Quadrant I, we can use the Pythagorean identity: \[ \sin^2 A + \cos^2 A = 1 \] Substituting \(\cos A\): \[ \sin^2 A + \left(\frac{4}{5}\right)^2 = 1 \] \[ \sin^2 A + \frac{16}{25} = 1 \] \[ \sin^2 A = 1 - \frac{16}{25} = \frac{9}{25} \] \[ \sin A = \sqrt{\frac{9}{25}} = \frac{3}{5} \quad (\text{since } A \text{ is in Quadrant I, } \sin A > 0) \] Now, we can find \(\tan A\): \[ \tan A = \frac{\sin A}{\cos A} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} \] Finally, we calculate \(\sin A \cdot \tan A\): \[ \sin A \cdot \tan A = \frac{3}{5} \cdot \frac{3}{4} = \frac{9}{20} \] Thus, the answer is \( \boxed{\frac{9}{20}} \).