The acute angle between the pair of tangents drawn to the ellipse $$2x^2 + 3y^2 = 5$$ from the point $$(1, 3)$$ is

We need to find the acute angle between the pair of tangents drawn to the ellipse $$2x^2 + 3y^2 = 5$$ from the point $$(1, 3)$$.

$$\frac{x^2}{5/2} + \frac{y^2}{5/3} = 1$$

Here $$a^2 = \frac{5}{2}$$ and $$b^2 = \frac{5}{3}$$.

The tangent to the ellipse $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ with slope $$m$$ is:

$$y = mx \pm \sqrt{a^2m^2 + b^2}$$

$$y = mx \pm \sqrt{\frac{5m^2}{2} + \frac{5}{3}}$$

$$3 = m \pm \sqrt{\frac{5m^2}{2} + \frac{5}{3}}$$

$$(3 - m)^2 = \frac{5m^2}{2} + \frac{5}{3}$$

$$9 - 6m + m^2 = \frac{5m^2}{2} + \frac{5}{3}$$

Multiply through by 6:

$$54 - 36m + 6m^2 = 15m^2 + 10$$

$$9m^2 + 36m - 44 = 0$$

$$m_1 + m_2 = -\frac{36}{9} = -4$$

$$m_1 m_2 = -\frac{44}{9}$$

$$\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|$$

First, find $$|m_1 - m_2|$$:

$$(m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1m_2 = 16 + \frac{176}{9} = \frac{144 + 176}{9} = \frac{320}{9}$$

$$|m_1 - m_2| = \frac{\sqrt{320}}{3} = \frac{8\sqrt{5}}{3}$$

Next, find $$1 + m_1m_2$$:

$$1 + m_1m_2 = 1 - \frac{44}{9} = -\frac{35}{9}$$

Therefore:

$$\tan\theta = \left|\frac{8\sqrt{5}/3}{-35/9}\right| = \frac{8\sqrt{5}}{3} \times \frac{9}{35} = \frac{72\sqrt{5}}{105} = \frac{24\sqrt{5}}{35} = \frac{24}{7\sqrt{5}}$$

So $$\theta = \tan^{-1}\left(\frac{24}{7\sqrt{5}}\right)$$.

Therefore, the correct answer is Option B: $$\tan^{-1}\dfrac{24}{7\sqrt{5}}$$.