The figure is not drawn to scale. EGHK is a parallelogram, DEK and EFK are isosceles triangles. KH//LT and ∠HST= 104°. DG is straight line.
- Find ∠DKE.
- Find ∠GFK.
(a)
∠HST = 104°
∠KPR = 104° (Corresponding angles)
∠NPK
= 180° - 104°
= 76° (Angles on a straight line)
∠DKE
= 180° - 76° - 76°
= 28° (Isosceles triangle)
(b)
∠NPK = ∠DEK = 76° (Corresponding angles)
∠KEF
= 180° - 76°
= 104° (Angles on a straight line)
∠EFR
= (180° - 104°) ÷ 2
= 38° (Isosceles triangle)
∠GFK
= 180° - 38°
= 142° (Angles on a straight line)
Answer(s): (a) 28°; (b) 142°