One through eight, i really need , it's geometry

Answers:

1) ∠U = 70°

2) ∠U = 65°

3) ∠TUY = 120°

4) ∠SRP = 105°

5) ∠E = 100°

6) ∠U = 30°

7) ∠FGT = 86°

8) ∠QPG = 130°

Explanation:

All given problem tackle the concept of the exterior angle theorem

This theorem states that:

"In any given triangle, the measure of any of its exterior angles is equal to the summation of its two non-adjacent interior angles"

Now, let's apply this to the given problems:

Problem 1:

120° is the exterior angle and ∠U and ∠T are the two non-adjacent interior angles

Therefore:

120° = ∠U + ∠T

120° = ∠U + 50°

∠U = 70°

Problem 2:

115° is the exterior angle and ∠U and ∠V are the two non-adjacent interior angles

Therefore:

115° = ∠U + ∠V

115° = ∠U + 50°

∠U = 65°

Problem 3:

∠TUY is the exterior angle and ∠S and ∠T are the two non-adjacent interior angles

Therefore:

∠TUY = ∠S + ∠T

∠TUY = 70° + 50°

∠TUY = 120°

Problem 4:

∠SRP is the exterior angle and ∠S and ∠T are the two non-adjacent interior angles

Therefore:

∠SRP = ∠S + ∠T

∠SRP = 25° + 80°

∠SRP = 105°

Problem 5:

140° is the exterior angle and ∠E and ∠D are the two non-adjacent interior angles

Therefore:

140° = ∠E + ∠D

140° = ∠E + 40°

∠E = 100°

Problem 6:

110° is the exterior angle and ∠T and ∠U are the two non-adjacent interior angles

Therefore:

110° = ∠U + ∠T

110° = ∠U + 80°

∠U = 30°

Problem 7:

∠FGT is the exterior angle and ∠E and ∠F are the two non-adjacent interior angles

Therefore:

∠FGT = ∠E + ∠F

∠FGT = 28° + 58°

∠FGT = 86°

Problem 8:

∠QPG is the exterior angle and ∠Q and ∠R are the two non-adjacent interior angles

Therefore:

∠QPG = ∠Q + ∠R

∠QPG = 95° + 35°

∠QPG = 130°

Hope this helps :)

it's both, now can you answer my most recent question?

i believe the correct answer is b.