If $t_a,\ t_b,\ t_c$ are medians of triangle, $\alpha,\ \beta,\ \gamma$ interior angles, $P$ area of triangle, prove formulas:
$${t_a}^2=\frac{1}{4}a^2+2P\cot\alpha$$
$${t_b}^2=\frac{1}{4}b^2+2P\cot\beta$$
$${t_c}^2=\frac{1}{4}c^2+2P\cot\gamma$$

Yes, this is solution. You forgot to mention formulas $abc=4PR$ and $\frac{a}{\sin\alpha}=2R$, where $P$ is area of triangle, and $R$ is radius of circumscribed circle.