The figure is not drawn to scale. BDEF is a parallelogram, ABF and BCF are isosceles triangles. FE//GP and ∠ENP= 100°. AD is straight line.
- Find ∠AFB.
- Find ∠DCF.
(a)
∠ENP = 100°
∠FKL = 100° (Corresponding angles)
∠HKF
= 180° - 100°
= 80° (Angles on a straight line)
∠AFB
= 180° - 80° - 80°
= 20° (Isosceles triangle)
(b)
∠HKF = ∠ABF = 80° (Corresponding angles)
∠FBC
= 180° - 80°
= 100° (Angles on a straight line)
∠BCL
= (180° - 100°) ÷ 2
= 40° (Isosceles triangle)
∠DCF
= 180° - 40°
= 140° (Angles on a straight line)
Answer(s): (a) 20°; (b) 140°
