PSLEIn the figure, ABCD is a straight line, BDEF is a parallelogram and EC = EF. ∠FAB is a right angle, ∠AFB = 28° and ∠DEC = 14°.
- Find ∠g.
- Find ∠h.
(a)
∠FAB = 90° (Right angle)
∠g
= 90° + 28°
= 118° (Exterior angle of a triangle)
(b)
∠DEF = ∠DBF = 118° (Parallelogram)
∠CEF
= 118° - 14°
= 104°
∠h
= (180° - 104°) ÷ 2
= 38° (Isosceles triangle)
Answer(s): (a) 118°; (b) 38°