En ambos casos faltan 2 lados y se tiene 2 angulos
Con teorema del seno se sacan
[tex]\frac{9,8cm}{sen(71^o)} = \frac{AC}{sen(51^o)}\\\\\frac{9,8cm}{sen(71^o)} = \frac{AB}{sen(180^o-71^o-51^o)}\\\\\frac{50cm}{sen(84^o)} = \frac{KL}{sen(42^o)}\\\\\frac{50cm}{sen(84^o)} = \frac{JK}{sen(180^o-84^o-42^o)}[/tex]
