PSLEIn the figure, BCDE is a straight line, CEFG is a parallelogram and FD = FG. ∠GBC is a right angle, ∠BGC = 28° and ∠EFD = 14°.
- Find ∠h.
- Find ∠j.
(a)
∠GBC = 90° (Right angle)
∠h
= 90° + 28°
= 118° (Exterior angle of a triangle)
(b)
∠EFG = ∠ECG = 118° (Parallelogram)
∠DFG
= 118° - 14°
= 104°
∠j
= (180° - 104°) ÷ 2
= 38° (Isosceles triangle)
Answer(s): (a) 118°; (b) 38°