The figure is not drawn to scale. EGHK is a parallelogram, DEK and EFK are isosceles triangles. KH//LT and ∠HST= 108°. DG is straight line.
- Find ∠DKE.
- Find ∠GFK.
(a)
∠HST = 108°
∠KPR = 108° (Corresponding angles)
∠NPK
= 180° - 108°
= 72° (Angles on a straight line)
∠DKE
= 180° - 72° - 72°
= 36° (Isosceles triangle)
(b)
∠NPK = ∠DEK = 72° (Corresponding angles)
∠KEF
= 180° - 72°
= 108° (Angles on a straight line)
∠EFR
= (180° - 108°) ÷ 2
= 36° (Isosceles triangle)
∠GFK
= 180° - 36°
= 144° (Angles on a straight line)
Answer(s): (a) 36°; (b) 144°