In the figure, CDEF is a parallelogram. O is the centre of the circle. Find
- ∠CFE
- ∠CDF
(a)
∠ODF = ∠DFO = 28° (Angle properties within a circle)
∠FOD
= 180° - 28° - 28°
= 124° (Angles sum of triangle)
∠OCD + ∠ODC = ∠DOF (Exterior angle of a triangle)
∠OCD = ∠ODC (Isosceles triangle, OC = OD)
∠OCD
= 124° ÷ 2
= 62° (Exterior angle of a triangle)
∠CFE
= 180° - 62°
= 118° (Interior angles)
(b)
∠CDF
= 180° - 62° - 28°
= 90° (Angle sum of triangles)
Answer(s): (a) 118°; (b) 90°