Answer:
[tex]x=60^o[/tex]
Step-by-step explanation:
The picture of the question in the attached figure
step 1
Find the measure of angle BEG
we know that
[tex]m\angle BED+m\angle BEG+m\angle GEF=180^o[/tex] ----> by supplementary angles (form a linear pair)
substitute the given values
[tex]110^o+m\angle BEG+25^o=180^o[/tex]
[tex]m\angle BEG=180^o-135^o=45^o[/tex]
step 2
Find the measure of angle HBC
we know that
[tex]m\angle HBC+m\angle ABH=180^o[/tex] ---> by supplementary angles (form a linear pair)
we know that
[tex]m\angle ABH=110^o[/tex] ---> by corresponding angles
substitute
[tex]m\angle HBC+110^0=180^o[/tex]
[tex]m\angle HBC=180^o-110^o=70^o[/tex]
step 3
Find the measure of angle EBG
we know that
[tex]m\angle EBG+m\angle GBC+m\ angle HBC=180^o[/tex]
substitute the given values
[tex]m\angle EBG+35^o+70^o=180^o[/tex]
[tex]m\angle EBG=180^o-105^o=75^o[/tex]
step 4
Find the measure of angle x
we know that
The sum of the interior angles in any triangle must be equal to 180 degrees
so
In the triangle BEG
[tex]m\angle BEG+m\angle BGE+m\angle EBG=180^o[/tex]
substitute the given values
[tex]45^o+x+75^o=180^o[/tex]
[tex]x=180^o-120^o=60^o[/tex]
