Hmm 5, 12, 13, this seems familiar

$tan\frac { A }{ 2 } tan\frac { B }{ 2 } \quad +\quad tan\frac { B }{ 2 } tan\frac { C }{ 2 } \quad +\quad tan\frac { C }{ 2 } tan\frac { A }{ 2 } \quad =\quad 1\\ tan\frac { A }{ 2 } \quad =\quad x\quad ;\quad tan\frac { B }{ 2 } \quad =\quad y\quad ;\quad tan\frac { C }{ 2 } \quad =\quad z\quad ;\\ with\quad A,B,C\quad is\quad angle\quad in\quad triangle.\\ 5(tan\frac { A }{ 2 } \quad +\quad cot\frac { A }{ 2 } )\quad =\quad 12(tan\frac { B }{ 2 } \quad +\quad cot\frac { B }{ 2 } )\quad =13(tan\frac { C }{ 2 } \quad +\quad cot\frac { C }{ 2 } )\\ \Longleftrightarrow \frac { 10 }{ sinA } =\frac { 24 }{ sinB } =\frac { 26 }{ sinC } \\ { 10 }^{ 2 }\quad +\quad { 24 }^{ 2 }\quad =\quad { 26 }^{ 2 }\quad \Longrightarrow \quad C\quad =\quad \frac { \pi }{ 2 } \quad \\ \Longrightarrow tanA\quad =\quad \frac { 10 }{ 24 } \quad \\ \Longrightarrow tan\frac { A }{ 2 } \quad =\quad \boxed{0.2}$