Answer:
[tex]x=105[/tex]°
Step-by-step explanation:
Given:
Let us label the vertices of the figure as shown below.
Consider triangle CDE. The exterior angle CDB is equal to the sum of opposite interior angles. Therefore,
[tex]\angle CDB=\angle CED+\angle DCE\\\angle CDB=80+50=130\°[/tex]
Now, for a quadrilateral ABDX, the exterior angle DXC is given as:
[tex]\angle DXC=\angle BAX+\angle ABD+\angle XDB-180\\x=65+90+130-180\\x=285-180=105\°[/tex]
Therefore, [tex]x=105\°[/tex]
