The figure is not drawn to scale. EGHK is a parallelogram, DEK and EFK are isosceles triangles. KH//LT and ∠HST= 114°. DG is straight line.
- Find ∠DKE.
- Find ∠GFK.
(a)
∠HST = 114°
∠KPR = 114° (Corresponding angles)
∠NPK
= 180° - 114°
= 66° (Angles on a straight line)
∠DKE
= 180° - 66° - 66°
= 48° (Isosceles triangle)
(b)
∠NPK = ∠DEK = 66° (Corresponding angles)
∠KEF
= 180° - 66°
= 114° (Angles on a straight line)
∠EFR
= (180° - 114°) ÷ 2
= 33° (Isosceles triangle)
∠GFK
= 180° - 33°
= 147° (Angles on a straight line)
Answer(s): (a) 48°; (b) 147°