The figure is not drawn to scale. BDEF is a parallelogram, ABF and BCF are isosceles triangles. FE//GP and ∠ENP= 120°. AD is straight line.
- Find ∠AFB.
- Find ∠DCF.
(a)
∠ENP = 120°
∠FKL = 120° (Corresponding angles)
∠HKF
= 180° - 120°
= 60° (Angles on a straight line)
∠AFB
= 180° - 60° - 60°
= 60° (Isosceles triangle)
(b)
∠HKF = ∠ABF = 60° (Corresponding angles)
∠FBC
= 180° - 60°
= 120° (Angles on a straight line)
∠BCL
= (180° - 120°) ÷ 2
= 30° (Isosceles triangle)
∠DCF
= 180° - 30°
= 150° (Angles on a straight line)
Answer(s): (a) 60°; (b) 150°