PSLEIn the figure, BCDE is a straight line, CEFG is a parallelogram and FD = FG. ∠GBC is a right angle, ∠BGC = 36° and ∠EFD = 12°.
- Find ∠h.
- Find ∠j.
(a)
∠GBC = 90° (Right angle)
∠h
= 90° + 36°
= 126° (Exterior angle of a triangle)
(b)
∠EFG = ∠ECG = 126° (Parallelogram)
∠DFG
= 126° - 12°
= 114°
∠j
= (180° - 114°) ÷ 2
= 33° (Isosceles triangle)
Answer(s): (a) 126°; (b) 33°