PSLEIn the figure, BCDE is a straight line, CEFG is a parallelogram and FD = FG. ∠GBC is a right angle, ∠BGC = 31° and ∠EFD = 21°.
- Find ∠h.
- Find ∠j.
(a)
∠GBC = 90° (Right angle)
∠h
= 90° + 31°
= 121° (Exterior angle of a triangle)
(b)
∠EFG = ∠ECG = 121° (Parallelogram)
∠DFG
= 121° - 21°
= 100°
∠j
= (180° - 100°) ÷ 2
= 40° (Isosceles triangle)
Answer(s): (a) 121°; (b) 40°