In the figure, BCDE is a quadrilateral where ∠BCD = 127° and ∠CDE = 110°. ∠DEF = 80° and CD = DE. The point F on BE is such that CF is parallel to DE. Calculate
- ∠CEF
- ∠CBF
(a)
∠DEC
= (180° - 110°) ÷ 2
= 35° (Isosceles triangle)
∠CEF
= 80° - 35°
= 45°
(b)
∠CBF
= 360° - 110° - 80° - 127°
= 43° (Sum of angles in a quadrilateral)
Answer(s): (a) 45°; (b) 43°