PSLEIn the figure, DEFG is a straight line, EGHJ is a parallelogram and HF = HJ. ∠JDE is a right angle, ∠DJE = 28° and ∠GHF = 12°.
- Find ∠k.
- Find ∠l.
(a)
∠JDE = 90° (Right angle)
∠k
= 90° + 28°
= 118° (Exterior angle of a triangle)
(b)
∠GHJ = ∠GEJ = 118° (Parallelogram)
∠FHJ
= 118° - 12°
= 106°
∠l
= (180° - 106°) ÷ 2
= 37° (Isosceles triangle)
Answer(s): (a) 118°; (b) 37°