PSLEIn the figure, CDEF is a straight line, DFGH is a parallelogram and GE = GH. ∠HCD is a right angle, ∠CHD = 27° and ∠FGE = 17°.
- Find ∠j.
- Find ∠k.
(a)
∠HCD = 90° (Right angle)
∠j
= 90° + 27°
= 117° (Exterior angle of a triangle)
(b)
∠FGH = ∠FDH = 117° (Parallelogram)
∠EGH
= 117° - 17°
= 100°
∠k
= (180° - 100°) ÷ 2
= 40° (Isosceles triangle)
Answer(s): (a) 117°; (b) 40°