If \( \sin A=\frac{2 \sqrt{6}}{5} \) and \( A \in\left[90^{\circ} ; 360^{\circ}\right] \) calculate without the use of a calculator and with the aid of a diagram the value of \( 15 \tan A \cdot \cos A \) If \( \sin \alpha=\frac{3}{5} \) with \( \alpha \in\left[00^{\circ}\right. \)

To find \( 15 \tan A \cdot \cos A \), we first need to determine \( \tan A \) and \( \cos A \) given that \( \sin A = \frac{2 \sqrt{6}}{5} \). Using the Pythagorean identity, \( \cos^2 A = 1 - \sin^2 A \): \[ \sin^2 A = \left(\frac{2 \sqrt{6}}{5}\right)^2 = \frac{24}{25} \] \[ \cos^2 A = 1 - \frac{24}{25} = \frac{1}{25} \implies \cos A = \pm \frac{1}{5} \] Since \( A \) can be in the third or fourth quadrant in the interval \([90^\circ, 360^\circ]\), and where \( \cos A\) is negative in the third quadrant: \[ \cos A = -\frac{1}{5} \] Next, we can find \( \tan A \): \[ \tan A = \frac{\sin A}{\cos A} = \frac{\frac{2 \sqrt{6}}{5}}{-\frac{1}{5}} = -2 \sqrt{6} \] Now compute \( 15 \tan A \cdot \cos A\): \[ 15 \tan A \cdot \cos A = 15 \cdot (-2 \sqrt{6}) \cdot \left(-\frac{1}{5}\right) = 15 \cdot \frac{2 \sqrt{6}}{5} = 6 \sqrt{6} \] Thus, the final result is: \[ \boxed{6 \sqrt{6}} \] Now regarding \( \sin \alpha = \frac{3}{5} \), to find \( \cos \alpha \): \[ \cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25} \implies \cos \alpha = \pm \frac{4}{5} \] And since \( \alpha \) is within \([0^\circ, 90^\circ]\), where \( \cos \alpha \) is positive: \[ \cos \alpha = \frac{4}{5} \]