The figure is not drawn to scale. ABE is an equilateral triangle and DBC is an isosceles triangle. CBA and BDF are straight lines and CA//FE. ∠EBF = 90° and ∠DEB = 55°.
- Find ∠CBF.
- Find ∠BDC.
- Find ∠EDF.
(a)
∠EBA = 60° (Equilateral triangle ABY)
∠EBF = 90°
∠DEB = 55°
∠CBF
= 180° - ∠EBF - ∠EBA
= 180° - 90° - 60°
= 30° (Angles on a straight line)
(b)
∠BDC
= (180° - ∠CBF ÷ 2
= (180° - 30°) ÷ 2
= 150° ÷ 2
= 75° (Isosceles triangle)
(c)
∠EDB
= 180° - ∠EBF - ∠DEB
= 180° - 90° - 55°
= 35°
∠EDF
= 180° - ∠EDB
= 180° - 35°
= 145° (Angles on a straight line)
Answer(s): (a) 30°; (b) 75°; (c) 145°
