The figure is not drawn to scale. EGHK is a parallelogram, DEK and EFK are isosceles triangles. KH//LT and ∠HST= 118°. DG is straight line.
- Find ∠DKE.
- Find ∠GFK.
(a)
∠HST = 118°
∠KPR = 118° (Corresponding angles)
∠NPK
= 180° - 118°
= 62° (Angles on a straight line)
∠DKE
= 180° - 62° - 62°
= 56° (Isosceles triangle)
(b)
∠NPK = ∠DEK = 62° (Corresponding angles)
∠KEF
= 180° - 62°
= 118° (Angles on a straight line)
∠EFR
= (180° - 118°) ÷ 2
= 31° (Isosceles triangle)
∠GFK
= 180° - 31°
= 149° (Angles on a straight line)
Answer(s): (a) 56°; (b) 149°