Respuesta :

gmany

[tex]m\angle DCA=m\angle BCA\ \text{therefore:}\\\\4x=6x-58\ \ \ \ |-6x\\\\-2x=-58\ \ \ |:(-2)\\\\x=29\\\\m\angle DCA=(4x)^o\to m\angle DCA=(4\cdot29)^o=116^o[/tex]

Answer: C. 116