If θ is the angle between the pair of tangents drawn to the ellipse...

The equation of the tangent to $3{x}^{2} + 2{y}^{2} = 5$ in slope form is $y = mx \pm \sqrt{\dfrac{5}{3}{m}^{2} + \dfrac{5}{2}}$

If it passes through \left(1,2\right)$, then</p><p>${\left(2 - m\right)}^{2} = \dfrac{5}{3}{m}^{2} + \dfrac{5}{2}$<br /><br />$⇒ \dfrac{2{m}^{2}}{3} + 4m - \dfrac{3}{2} = 0$<br /><br />$⇒ {m}_{1} + {m}_{2} = - \dfrac{4}{\dfrac{2}{3}} \text{$and$} {m}_{1}{m}_{2} = - \dfrac{\dfrac{3}{2}}{\dfrac{2}{3}}$<br /><br />$⇒ {m}_{1} + {m}_{2} = - 6 \text{$and$} {m}_{1}{m}_{2} = - \dfrac{9}{4}$<br /><br />\left|\tan \theta\right| = \dfrac{\sqrt{{\left({m}_{1} + {m}_{2}\right)}^{2} - 4{m}_{1}{m}_{2}}}{\left[1 + {m}_{1}{m}_{2}\right]}$

$ta{n}^{2}\theta = \dfrac{36 + 9}{{\left(- \dfrac{5}{4}\right)}^{2}} = \dfrac{45 \times 16}{25} = \dfrac{144}{5} = 28.8$