PSLEIn the figure, DEFG is a straight line, EGHJ is a parallelogram and HF = HJ. ∠JDE is a right angle, ∠DJE = 34° and ∠GHF = 12°.
- Find ∠k.
- Find ∠l.
(a)
∠JDE = 90° (Right angle)
∠k
= 90° + 34°
= 124° (Exterior angle of a triangle)
(b)
∠GHJ = ∠GEJ = 124° (Parallelogram)
∠FHJ
= 124° - 12°
= 112°
∠l
= (180° - 112°) ÷ 2
= 34° (Isosceles triangle)
Answer(s): (a) 124°; (b) 34°