PSLEIn the figure, ABCD is a straight line, BDEF is a parallelogram and EC = EF. ∠FAB is a right angle, ∠AFB = 23° and ∠DEC = 17°.
- Find ∠g.
- Find ∠h.
(a)
∠FAB = 90° (Right angle)
∠g
= 90° + 23°
= 113° (Exterior angle of a triangle)
(b)
∠DEF = ∠DBF = 113° (Parallelogram)
∠CEF
= 113° - 17°
= 96°
∠h
= (180° - 96°) ÷ 2
= 42° (Isosceles triangle)
Answer(s): (a) 113°; (b) 42°