The figure is not drawn to scale. EGHK is a parallelogram, DEK and EFK are isosceles triangles. KH//LT and ∠HST= 113°. DG is straight line.
- Find ∠DKE.
- Find ∠GFK.
(a)
∠HST = 113°
∠KPR = 113° (Corresponding angles)
∠NPK
= 180° - 113°
= 67° (Angles on a straight line)
∠DKE
= 180° - 67° - 67°
= 46° (Isosceles triangle)
(b)
∠NPK = ∠DEK = 67° (Corresponding angles)
∠KEF
= 180° - 67°
= 113° (Angles on a straight line)
∠EFR
= (180° - 113°) ÷ 2
= 33.5° (Isosceles triangle)
∠GFK
= 180° - 33.5°
= 146.5° (Angles on a straight line)
Answer(s): (a) 46°; (b) 146.5°