PSLEIn the figure, CDEF is a straight line, DFGH is a parallelogram and GE = GH. ∠HCD is a right angle, ∠CHD = 38° and ∠FGE = 14°.
- Find ∠j.
- Find ∠k.
(a)
∠HCD = 90° (Right angle)
∠j
= 90° + 38°
= 128° (Exterior angle of a triangle)
(b)
∠FGH = ∠FDH = 128° (Parallelogram)
∠EGH
= 128° - 14°
= 114°
∠k
= (180° - 114°) ÷ 2
= 33° (Isosceles triangle)
Answer(s): (a) 128°; (b) 33°