PSLEIn the figure, DEFG is a straight line, EGHJ is a parallelogram and HF = HJ. ∠JDE is a right angle, ∠DJE = 38° and ∠GHF = 14°.
- Find ∠k.
- Find ∠l.
(a)
∠JDE = 90° (Right angle)
∠k
= 90° + 38°
= 128° (Exterior angle of a triangle)
(b)
∠GHJ = ∠GEJ = 128° (Parallelogram)
∠FHJ
= 128° - 14°
= 114°
∠l
= (180° - 114°) ÷ 2
= 33° (Isosceles triangle)
Answer(s): (a) 128°; (b) 33°