In the figure, PQR and RST are isosceles triangles where PR = QR and RS = RT. Given that SU, TW and PY are straight lines, find
- ∠VZY
- ∠TUZ
(a)
∠PQR = ∠QPR = ∠RST = ∠RTS (Isosceles triangle)
∠RTS = ∠UTZ = 47° (Verticallly opposite angles)
∠WTZ
= 47° - 13°
= 34°
∠TVZ
= 180° - 95°
= 85° (Angles on a straight line)
∠VZY
= 85° + 34°
= 119° (Exterior angle of a triangle)
(b)
∠TUZ
= 119° - 47°
= 72° (Exterior angle of a triangle)
Answer(s): (a) 119°; (b) 72°