The figure is not drawn to scale. EGHK is a parallelogram, DEK and EFK are isosceles triangles. KH//LT and ∠HST= 101°. DG is straight line.
- Find ∠DKE.
- Find ∠GFK.
(a)
∠HST = 101°
∠KPR = 101° (Corresponding angles)
∠NPK
= 180° - 101°
= 79° (Angles on a straight line)
∠DKE
= 180° - 79° - 79°
= 22° (Isosceles triangle)
(b)
∠NPK = ∠DEK = 79° (Corresponding angles)
∠KEF
= 180° - 79°
= 101° (Angles on a straight line)
∠EFR
= (180° - 101°) ÷ 2
= 39.5° (Isosceles triangle)
∠GFK
= 180° - 39.5°
= 140.5° (Angles on a straight line)
Answer(s): (a) 22°; (b) 140.5°
