The figure is not drawn to scale. EGHK is a parallelogram, DEK and EFK are isosceles triangles. KH//LT and ∠HST= 120°. DG is straight line.
- Find ∠DKE.
- Find ∠GFK.
(a)
∠HST = 120°
∠KPR = 120° (Corresponding angles)
∠NPK
= 180° - 120°
= 60° (Angles on a straight line)
∠DKE
= 180° - 60° - 60°
= 60° (Isosceles triangle)
(b)
∠NPK = ∠DEK = 60° (Corresponding angles)
∠KEF
= 180° - 60°
= 120° (Angles on a straight line)
∠EFR
= (180° - 120°) ÷ 2
= 30° (Isosceles triangle)
∠GFK
= 180° - 30°
= 150° (Angles on a straight line)
Answer(s): (a) 60°; (b) 150°