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If $$\cos \mathrm{A}=\frac{2 \sqrt{6}}{5}$$ and $$\mathrm{A} \in\left[90^{\circ} ; 360^{\circ}\right]$$ calculate without the use of a calculator and with the aid of a diagram the value of $$5 \tan A \cdot \cos \mathrm{A}$$
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Given that $$\cos A = \frac{2\sqrt{6}}{5}$$ and $$A$$ is in the interval $$[90^\circ, 360^\circ]$$, which implies $$A$$ is in the fourth quadrant where $$\cos A > 0$$ and $$\sin A < 0$$. We use the Pythagorean identity to find $$\sin A$$:

$\sin^2 A = 1 - \cos^2 A = 1 - \left(\frac{2\sqrt{6}}{5}\right)^2$

Calculating $$\sin^2 A$$ gives:

$\sin^2 A = 1 - \left(\frac{2\sqrt{6}}{5}\right)^2 = 1 - \frac{24}{25} = \frac{1}{25}$

Since $$A$$ is in the fourth quadrant, $$\sin A$$ is negative, thus:

$\sin A = -\sqrt{\frac{1}{25}} = -\frac{1}{5} = -0.2$

The tangent of $$A$$ can be calculated as $$\tan A = \frac{\sin A}{\cos A}$$, and thus:

$\tan A = \frac{-0.2}{\frac{2\sqrt{6}}{5}} = -\frac{1}{2\sqrt{6}} = -0.20412414523193162$

Finally, the value of $$5 \tan A \cdot \cos A$$ is calculated as:

$5 \tan A \cdot \cos A = 5 \cdot (-0.20412414523193162) \cdot \frac{2\sqrt{6}}{5} = -1.0$

Therefore, $$5 \tan A \cdot \cos A = -1.0$$.
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