SInce $5\cos{\beta}-3=0$ then

$$\cos{\beta} = \dfrac{3}{5}$$

$$\beta = \cos^{-1}{\left(\frac{3}{5}\right)}$$

Given that $\alpha + \beta = 90^{\circ}$, then:

\[ \alpha + \cos^{-1}{\left(\frac{3}{5}\right)} = 90^{\circ}\]

\[ \therefore \alpha = 90^{\circ} - \cos^{-1}{\left(\frac{3}{5}\right)} \]

$\cot{ \alpha} = \dfrac{1}{\tan{\alpha}}$

$$= \dfrac{1}{\tan{\left[90^{\circ} - \cos^{-1}{\left(\frac{3}{5}\right)}\right]}}$$

$\therefore \alpha = \dfrac{1}{\tan{\left[90^{\circ} - 53.1^{\circ}\right]}}$

$\therefore \alpha = \dfrac{1}{\tan{36.9^{\circ}}} = -0.9729$