The figure is not drawn to scale. EGHK is a parallelogram, DEK and EFK are isosceles triangles. KH//LT and ∠HST= 106°. DG is straight line.
- Find ∠DKE.
- Find ∠GFK.
(a)
∠HST = 106°
∠KPR = 106° (Corresponding angles)
∠NPK
= 180° - 106°
= 74° (Angles on a straight line)
∠DKE
= 180° - 74° - 74°
= 32° (Isosceles triangle)
(b)
∠NPK = ∠DEK = 74° (Corresponding angles)
∠KEF
= 180° - 74°
= 106° (Angles on a straight line)
∠EFR
= (180° - 106°) ÷ 2
= 37° (Isosceles triangle)
∠GFK
= 180° - 37°
= 143° (Angles on a straight line)
Answer(s): (a) 32°; (b) 143°