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

If it passes through $\left(\mathrm{1,2}\right)$, then

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

$\Rightarrow \frac{2m^2}{3}+4m-\frac{3}{2}=0$

$\Rightarrow m_1+m_2=-\frac{4}{\frac{2}{3}}\mathrm{ }\mathrm{ }\mathrm{ }\&\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }{m}_1{m}_2=-\frac{\frac{3}{2}}{\frac{2}{3}}$

$\Rightarrow m_1+m_2=-6\mathrm{ }\mathrm{ }\mathrm{ }\&\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }{m}_1{m}_2=-\frac{9}{4}$

$\left|\tan \theta \right|=\frac{\sqrt{{\left(m_1+m_2\right)}^2-4m_1{m}_2}}{\left[1+m_1{m}_2\right]}$

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