The figure is not drawn to scale. EGHK is a parallelogram, DEK and EFK are isosceles triangles. KH//LT and ∠HST= 119°. DG is straight line.
- Find ∠DKE.
- Find ∠GFK.
(a)
∠HST = 119°
∠KPR = 119° (Corresponding angles)
∠NPK
= 180° - 119°
= 61° (Angles on a straight line)
∠DKE
= 180° - 61° - 61°
= 58° (Isosceles triangle)
(b)
∠NPK = ∠DEK = 61° (Corresponding angles)
∠KEF
= 180° - 61°
= 119° (Angles on a straight line)
∠EFR
= (180° - 119°) ÷ 2
= 30.5° (Isosceles triangle)
∠GFK
= 180° - 30.5°
= 149.5° (Angles on a straight line)
Answer(s): (a) 58°; (b) 149.5°