The figure is not drawn to scale. EGHK is a parallelogram, DEK and EFK are isosceles triangles. KH//LT and ∠HST= 107°. DG is straight line.
- Find ∠DKE.
- Find ∠GFK.
(a)
∠HST = 107°
∠KPR = 107° (Corresponding angles)
∠NPK
= 180° - 107°
= 73° (Angles on a straight line)
∠DKE
= 180° - 73° - 73°
= 34° (Isosceles triangle)
(b)
∠NPK = ∠DEK = 73° (Corresponding angles)
∠KEF
= 180° - 73°
= 107° (Angles on a straight line)
∠EFR
= (180° - 107°) ÷ 2
= 36.5° (Isosceles triangle)
∠GFK
= 180° - 36.5°
= 143.5° (Angles on a straight line)
Answer(s): (a) 34°; (b) 143.5°