The figure is not drawn to scale. BDEF is a parallelogram, ABF and BCF are isosceles triangles. FE//GP and ∠ENP= 112°. AD is straight line.
- Find ∠AFB.
- Find ∠DCF.
(a)
∠ENP = 112°
∠FKL = 112° (Corresponding angles)
∠HKF
= 180° - 112°
= 68° (Angles on a straight line)
∠AFB
= 180° - 68° - 68°
= 44° (Isosceles triangle)
(b)
∠HKF = ∠ABF = 68° (Corresponding angles)
∠FBC
= 180° - 68°
= 112° (Angles on a straight line)
∠BCL
= (180° - 112°) ÷ 2
= 34° (Isosceles triangle)
∠DCF
= 180° - 34°
= 146° (Angles on a straight line)
Answer(s): (a) 44°; (b) 146°