ΔVUQ
∠V + ∠U + ∠Q = 180° triangle sum theorem
b + 75° + 75° = 180° substitution
b + 150° = 180° simplify (add like terms)
b = 30° subtraction property of equality
ΔUWT
∠U is a vertical angle so ∠U = 75°
∠Tand 101° is a linear pair so ∠T + 101° = 180° ⇒ ∠T = 79°
∠U + ∠W + ∠T = 180°
75° + e + 79 = 180°
e + 154° = 180°
e = 26°
ΔTXS
∠T and 101° is a linear pair so ∠T + 101° = 180° ⇒ ∠T = 79°
∠T + ∠X + ∠S = 180°
79° + f + 88 = 180°
f + 167° = 180°
f = 13°
ΔVTY
∠V + ∠T + ∠Y = 180°
∠V was solved above (b = 30°)
∠T is a vertical angle so = 101°
30° + 101° + c = 180°
131° + c = 180°
c = 49°
ΔWZS
∠W + ∠Z + ∠S = 180°
∠W was solved above (e = 26°)
∠S is a linear pair with 88° so ∠S + 88° = 180° ⇒ ∠S = 92°
26° + d + 92° = 180°
d + 118° = 180°
d = 62°
Answer: b=30°, c=49°, d=62°, e=26°, f=13°
