In the figure, BCDE is a quadrilateral where ∠BCD = 125° and ∠CDE = 118°. ∠DEF = 71° and CD = DE. The point F on BE is such that CF is parallel to DE. Calculate
- ∠CEF
- ∠CBF
(a)
∠DEC
= (180° - 118°) ÷ 2
= 31° (Isosceles triangle)
∠CEF
= 71° - 31°
= 40°
(b)
∠CBF
= 360° - 118° - 71° - 125°
= 46° (Sum of angles in a quadrilateral)
Answer(s): (a) 40°; (b) 46°