The figure is not drawn to scale. EGHK is a parallelogram, DEK and EFK are isosceles triangles. KH//LT and ∠HST= 116°. DG is straight line.
- Find ∠DKE.
- Find ∠GFK.
(a)
∠HST = 116°
∠KPR = 116° (Corresponding angles)
∠NPK
= 180° - 116°
= 64° (Angles on a straight line)
∠DKE
= 180° - 64° - 64°
= 52° (Isosceles triangle)
(b)
∠NPK = ∠DEK = 64° (Corresponding angles)
∠KEF
= 180° - 64°
= 116° (Angles on a straight line)
∠EFR
= (180° - 116°) ÷ 2
= 32° (Isosceles triangle)
∠GFK
= 180° - 32°
= 148° (Angles on a straight line)
Answer(s): (a) 52°; (b) 148°