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° - 146°) ÷ 2
= 34° ÷ 2
= 17° (Isosceles triangle)
∠BDE
= 60° - 17°
= 43° (Equilateral triangle)
(b)
∠BFE
= 60° - 17°
= 43° (Equilateral triangle)
∠BEF
= 180° - 43° - 17°
= 120° (Angles sum of triangle)
∠BED
= 360° - 146° - 120°
= 94° (Angles at a point)
∠CBD
= ∠BDE
= 43° (Alternate angles)
∠CBF
= 60° + 43°
= 103°
Answer(s): (a) 43°; (b) 94°; (c) 103°