The figure is not drawn to scale. EGHK is a parallelogram, DEK and EFK are isosceles triangles. KH//LT and ∠HST= 100°. DG is straight line.
- Find ∠DKE.
- Find ∠GFK.
(a)
∠HST = 100°
∠KPR = 100° (Corresponding angles)
∠NPK
= 180° - 100°
= 80° (Angles on a straight line)
∠DKE
= 180° - 80° - 80°
= 20° (Isosceles triangle)
(b)
∠NPK = ∠DEK = 80° (Corresponding angles)
∠KEF
= 180° - 80°
= 100° (Angles on a straight line)
∠EFR
= (180° - 100°) ÷ 2
= 40° (Isosceles triangle)
∠GFK
= 180° - 40°
= 140° (Angles on a straight line)
Answer(s): (a) 20°; (b) 140°