[tex]\alpha=180^o-118^o=62^o.\\\\\text{We know. The sum of the angles measures in the triangle is 180}^o.\\\text{Therefore we have the equation:}\\\\(6x-9)+(4x+7)+62=180\\\\6x-9+4x+7+62=180\\\\10x+60=180\qquad\text{subtract 60 from both sides}\\\\10x=120\qquad\text{divide both sides by 10}\\\\x=12\\\\m\angle K=(6x-9)^o\to m\angle K=(6\cdot12-9)^o\\\\m\angle K=(72-9)^o\\\\\boxed{m\angle K=63^o}[/tex]
