The figure is not drawn to scale. EGHK is a parallelogram, DEK and EFK are isosceles triangles. KH//LT and ∠HST= 111°. DG is straight line.
- Find ∠DKE.
- Find ∠GFK.
(a)
∠HST = 111°
∠KPR = 111° (Corresponding angles)
∠NPK
= 180° - 111°
= 69° (Angles on a straight line)
∠DKE
= 180° - 69° - 69°
= 42° (Isosceles triangle)
(b)
∠NPK = ∠DEK = 69° (Corresponding angles)
∠KEF
= 180° - 69°
= 111° (Angles on a straight line)
∠EFR
= (180° - 111°) ÷ 2
= 34.5° (Isosceles triangle)
∠GFK
= 180° - 34.5°
= 145.5° (Angles on a straight line)
Answer(s): (a) 42°; (b) 145.5°