First let's add a line from point C so that it is PERPENDICULAR to line AD
Also, note that, ∠EDB is 120° since it lies on a line with a 60° angle.



Since we already know 2 angles in ∆CED, we can see that the remaining angle, ∠ECD, is 30°, which means ∆CED is a 30-60-90 right triangle.
So, ∆CED is a 30-60-90 right triangle AND we know that the hypotenuse CD has length 2, we can conclude that side ED has length 1.



Now draw a line from E to B.
Since ED = EB (both have length 1), we can see that ∆EDB is an ISOSCELES TRIANGLE, which means the two remaining angles are each 30°



Next, ∠EAB is 150° since it shares a line with a 30° angle
Also, since we were originally told that ∠CBA is 45°, we can conclude that ∠ABE is 15°


Now focus on ∆CBE
Notice that this triangle is an ISOSCELES triangle, because ∠ECB = ∠EBC = 30°
This means that side EC = side EB



Now focus on ∆EAB
Since we already know two of the angles in this triangle (150° and 15°), we can conclude that = ∠EAB = 15°



Stay focused on ∆EAB
This triangle is an ISOSCELES triangle, because ∠EAB = ∠EBA = 15°
This means that side AE = side EB



Now focus on ∆ACE
Since CE = EA, this is an ISOSCELES triangle
Since one angle = 90°, the other two angles are each 45°


At this point, we can see that x = 45° + 30° = 75°
Answer: D

Cheers,
Brent