The figure is not drawn to scale. EGHK is a parallelogram, DEK and EFK are isosceles triangles. KH//LT and ∠HST= 112°. DG is straight line.
- Find ∠DKE.
- Find ∠GFK.
(a)
∠HST = 112°
∠KPR = 112° (Corresponding angles)
∠NPK
= 180° - 112°
= 68° (Angles on a straight line)
∠DKE
= 180° - 68° - 68°
= 44° (Isosceles triangle)
(b)
∠NPK = ∠DEK = 68° (Corresponding angles)
∠KEF
= 180° - 68°
= 112° (Angles on a straight line)
∠EFR
= (180° - 112°) ÷ 2
= 34° (Isosceles triangle)
∠GFK
= 180° - 34°
= 146° (Angles on a straight line)
Answer(s): (a) 44°; (b) 146°