Answer:
Correct: 2, 3, 4, 6
Step-by-step explanation:
ABCDE is a regular pentagon. The sum of the measures of all interior angles in the pentagon is
[tex](n-2)\cdot 180^{\circ}=(5-2)\cdot 180^{\circ}=540^{\circ},[/tex]
then the measure of each interior angles in the regular pentegon is
[tex]\dfrac{540^{\circ}}{5}=108^{\circ}[/tex]
Triangle ABe is isosceles triangle, so the adjacent to the bas BE angle has the measure
[tex]m\angle ABE=\dfrac{180^{\circ}-108^{\circ}}{2}=36^{\circ}[/tex]
Therefore,
[tex]m\angle CBE=m\angle ABC-m\angle ABE=108^{\circ}-36^{\circ}=72^{\circ}[/tex]
(option 2 is correct)
Since [tex]m\angle BCG=108^{\circ},[/tex] then
[tex]m\angle EGC=108^{\circ }+4x[/tex]
The sum of the measures of all interior angles in quadrilateral BCGE is
[tex](n-2)\cdot 180^{\circ}=(4-2)\cdot 180^{\circ}=360^{\circ},[/tex]
(option 6 is correct), then
[tex]m\angle CBE+m\angle BEG+m\angle EGC+m\angle BCG=360^{\circ}\\ \\72^{\circ}+19x+3^{\circ}+108^{\circ}+4x+108^{\circ}=360^{\circ}\\ \\23x=69^{\circ}\\ \\x=3[/tex]
(option 1 is false)
[tex]m\angle EGC=108^{\circ}+4\cdot 3^{\circ}=120^{\circ}[/tex]
(option 3 is correct)
[tex]m\angle GCB+m\angle EGC=108^{\circ}+72^{\circ}=180^{\circ}[/tex]
(option 4 is correct)
[tex]m\angle BEC+m\angle EGC=(19\cdot 3+3+108+4\cdot 3)^{\circ}=180^{\circ}[/tex]
(option 5 is false)
