Two Four Thirteen

Direct solution We can use this formula to solve the problem $\tan \left( A + B + C \right) = \frac{\tan \left( A\right) + \tan \left( B \right) +\tan \left( C\right) - \tan \left( A\right) \tan \left( B\right) \tan \left( C\right) }{1- \tan \left( A\right) \tan \left( B\right) - \tan \left( B\right) \tan \left( C\right) - \tan \left( C\right) \tan \left( A\right) } = \frac{19-104}{1-8-52-26} = 1$ $\text{so}: A + B +C =\mathring{45} , \mathring{225}$ But $Tan(A) = 2$ so $A> \mathring{45}$ $\text{so} : A +B + C =\mathring{225}$

$\begin{aligned} \tan{(A+B)} & = \frac{\tan{A}+\tan{B}}{1-\tan{A} \tan{B}} \\ & = \frac{2+4}{1-(2)(4)} = \color{#D61F06}{-}\frac{6}{7} \quad \quad \small \color{#D61F06}{[\Rightarrow A+B > 90^\circ]} \\ \tan{(A+B+C)} & = \frac{\tan{(A+B)}+\tan{C}}{1-\tan{(A+B)} \tan{c}} \\ & = \frac{-\frac{6}{7}+13}{1- \left( - \frac{6}{7} \right)(13)} = \frac{\frac{85}{7}}{\frac{85}{7}} = 1 \\ \Rightarrow A+B+C & = \boxed{225}^\circ \quad \quad \small \color{#D61F06}{[> 90^\circ]} \end{aligned}$