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A)Corresponding Angles Theorem; ∠AGF and ∠EHD are congruent
B)Alternate Exterior Angles Theorem; ∠EGB and ∠EHD are congruent
C)Corresponding Angles Theorem; ∠EGB and ∠EHD are congruent
D)Alternate Exterior Angles Theorem; ∠AGF and ∠EHD are congruent

Corresponding Angles Theorem; ∠AGF and ∠EHD are congruent

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Explanation:

∠EGB and ∠EHD are angles that are both on the right hand side of the transversal line EF. Also, these two angles are above their adjacent parallel line. In other words, ∠EGB is above line AB, while ∠EHD is above line CD.

So this all means ∠EGB and ∠EHD are corresponding angles. By the corresponding angles theorem, we can go from AB and CD being parallel to ∠EGB = ∠EHD.

Earlier it was shown that ∠AGF = ∠EGB. Combine this with ∠EGB = ∠EHD, and we can effectively chain the statements together to end up with ∠AGF = ∠EHD.

In terms of a math term, we used the transitive property of equality. This property says that if a = b and b = c, then a = c. You can think of three towns A, B, C. If you go from town A to town B, then town B to town C, then overall you went from town A to town C. We can write it as such which simplifies to

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47