The figure is not drawn to scale. BDEF is a parallelogram, ABF and BCF are isosceles triangles. FE//GP and ∠ENP= 104°. AD is straight line.
- Find ∠AFB.
- Find ∠DCF.
(a)
∠ENP = 104°
∠FKL = 104° (Corresponding angles)
∠HKF
= 180° - 104°
= 76° (Angles on a straight line)
∠AFB
= 180° - 76° - 76°
= 28° (Isosceles triangle)
(b)
∠HKF = ∠ABF = 76° (Corresponding angles)
∠FBC
= 180° - 76°
= 104° (Angles on a straight line)
∠BCL
= (180° - 104°) ÷ 2
= 38° (Isosceles triangle)
∠DCF
= 180° - 38°
= 142° (Angles on a straight line)
Answer(s): (a) 28°; (b) 142°