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 :)
