PSLEIn the figure, DEFG is a straight line, EGHJ is a parallelogram and HF = HJ. ∠JDE is a right angle, ∠DJE = 27° and ∠GHF = 19°.
- Find ∠k.
- Find ∠l.
(a)
∠JDE = 90° (Right angle)
∠k
= 90° + 27°
= 117° (Exterior angle of a triangle)
(b)
∠GHJ = ∠GEJ = 117° (Parallelogram)
∠FHJ
= 117° - 19°
= 98°
∠l
= (180° - 98°) ÷ 2
= 41° (Isosceles triangle)
Answer(s): (a) 117°; (b) 41°