Question

Pe latura AB a triunghiului isoscel ABC,cu [AB] congruent [AC] se construieste in exterior patratul ABMN.Aflati masura unghiului NCB
Va rog scrieti raspuns e urgent

Answer 1

   
[tex]\displaystyle\\ \sphericalangle BAC = x\\ \sphericalangle CAN = x + 90\\ AN = AC\\ \Longrightarrow~~\Delta ANC~\text{ este triunghi isoscel}\\\\ \sphericalangle ACN = \frac{180 - (x+90)}{2} =\frac{180 - x-90}{2} = \frac{90 - x }{2} \\\\ \sphericalangle ACB = \frac{180 - x}{2}\\\\ \ \textless \ NCB=\sphericalangle ACB-\sphericalangle ACN= \frac{180 - x}{2}- \frac{90 - x }{2} =\\\\ = \frac{180 - x -(90-x)}{2}=\frac{180 - x -90+x}{2}=\frac{180-90}{2}= \frac{90}{2}= \boxed{45^o} \\[/tex]