PSLEIn the figure, BCDE is a straight line, CEFG is a parallelogram and FD = FG. ∠GBC is a right angle, ∠BGC = 33° and ∠EFD = 17°.
- Find ∠h.
- Find ∠j.
(a)
∠GBC = 90° (Right angle)
∠h
= 90° + 33°
= 123° (Exterior angle of a triangle)
(b)
∠EFG = ∠ECG = 123° (Parallelogram)
∠DFG
= 123° - 17°
= 106°
∠j
= (180° - 106°) ÷ 2
= 37° (Isosceles triangle)
Answer(s): (a) 123°; (b) 37°