PSLEIn the figure, ABCD is a straight line, BDEF is a parallelogram and EC = EF. ∠FAB is a right angle, ∠AFB = 34° and ∠DEC = 16°.
- Find ∠g.
- Find ∠h.
(a)
∠FAB = 90° (Right angle)
∠g
= 90° + 34°
= 124° (Exterior angle of a triangle)
(b)
∠DEF = ∠DBF = 124° (Parallelogram)
∠CEF
= 124° - 16°
= 108°
∠h
= (180° - 108°) ÷ 2
= 36° (Isosceles triangle)
Answer(s): (a) 124°; (b) 36°