PSLEBCDE is a parallelogram, BDF is an equilateral triangle and DE = EF.
- Find ∠BDE.
- Find ∠DEB.
- Find ∠FBC.
(a)
DE = EF
Triangle DEF is an isosceles triangle.
∠EDF
= (180° - 132°) ÷ 2
= 48° ÷ 2
= 24° (Isosceles triangle)
∠BDE
= 60° - 24°
= 36° (Equilateral triangle)
(b)
∠BFE
= 60° - 24°
= 36° (Equilateral triangle)
∠BEF
= 180° - 36° - 24°
= 120° (Angles sum of triangle)
∠BED
= 360° - 132° - 120°
= 108° (Angles at a point)
∠CBD
= ∠BDE
= 36° (Alternate angles)
∠CBF
= 60° + 36°
= 96°
Answer(s): (a) 36°; (b) 108°; (c) 96°
