The figure is not drawn to scale. EGHK is a parallelogram, DEK and EFK are isosceles triangles. KH//LT and ∠HST= 103°. DG is straight line.
- Find ∠DKE.
- Find ∠GFK.
(a)
∠HST = 103°
∠KPR = 103° (Corresponding angles)
∠NPK
= 180° - 103°
= 77° (Angles on a straight line)
∠DKE
= 180° - 77° - 77°
= 26° (Isosceles triangle)
(b)
∠NPK = ∠DEK = 77° (Corresponding angles)
∠KEF
= 180° - 77°
= 103° (Angles on a straight line)
∠EFR
= (180° - 103°) ÷ 2
= 38.5° (Isosceles triangle)
∠GFK
= 180° - 38.5°
= 141.5° (Angles on a straight line)
Answer(s): (a) 26°; (b) 141.5°