The figure is not drawn to scale. BDEF is a parallelogram, ABF and BCF are isosceles triangles. FE//GP and ∠ENP= 106°. AD is straight line.
- Find ∠AFB.
- Find ∠DCF.
(a)
∠ENP = 106°
∠FKL = 106° (Corresponding angles)
∠HKF
= 180° - 106°
= 74° (Angles on a straight line)
∠AFB
= 180° - 74° - 74°
= 32° (Isosceles triangle)
(b)
∠HKF = ∠ABF = 74° (Corresponding angles)
∠FBC
= 180° - 74°
= 106° (Angles on a straight line)
∠BCL
= (180° - 106°) ÷ 2
= 37° (Isosceles triangle)
∠DCF
= 180° - 37°
= 143° (Angles on a straight line)
Answer(s): (a) 32°; (b) 143°