PSLEIn the figure, CDEF is a straight line, DFGH is a parallelogram and GE = GH. ∠HCD is a right angle, ∠CHD = 24° and ∠FGE = 18°.
- Find ∠j.
- Find ∠k.
(a)
∠HCD = 90° (Right angle)
∠j
= 90° + 24°
= 114° (Exterior angle of a triangle)
(b)
∠FGH = ∠FDH = 114° (Parallelogram)
∠EGH
= 114° - 18°
= 96°
∠k
= (180° - 96°) ÷ 2
= 42° (Isosceles triangle)
Answer(s): (a) 114°; (b) 42°