PSLEIn the figure, BCDE is a straight line, CEFG is a parallelogram and FD = FG. ∠GBC is a right angle, ∠BGC = 38° and ∠EFD = 22°.
- Find ∠h.
- Find ∠j.
(a)
∠GBC = 90° (Right angle)
∠h
= 90° + 38°
= 128° (Exterior angle of a triangle)
(b)
∠EFG = ∠ECG = 128° (Parallelogram)
∠DFG
= 128° - 22°
= 106°
∠j
= (180° - 106°) ÷ 2
= 37° (Isosceles triangle)
Answer(s): (a) 128°; (b) 37°