The figure is not drawn to scale. EGHK is a parallelogram, DEK and EFK are isosceles triangles. KH//LT and ∠HST= 109°. DG is straight line.
- Find ∠DKE.
- Find ∠GFK.
(a)
∠HST = 109°
∠KPR = 109° (Corresponding angles)
∠NPK
= 180° - 109°
= 71° (Angles on a straight line)
∠DKE
= 180° - 71° - 71°
= 38° (Isosceles triangle)
(b)
∠NPK = ∠DEK = 71° (Corresponding angles)
∠KEF
= 180° - 71°
= 109° (Angles on a straight line)
∠EFR
= (180° - 109°) ÷ 2
= 35.5° (Isosceles triangle)
∠GFK
= 180° - 35.5°
= 144.5° (Angles on a straight line)
Answer(s): (a) 38°; (b) 144.5°