The figure is not drawn to scale. BDEF is a parallelogram, ABF and BCF are isosceles triangles. FE//GP and ∠ENP= 116°. AD is straight line.
- Find ∠AFB.
- Find ∠DCF.
(a)
∠ENP = 116°
∠FKL = 116° (Corresponding angles)
∠HKF
= 180° - 116°
= 64° (Angles on a straight line)
∠AFB
= 180° - 64° - 64°
= 52° (Isosceles triangle)
(b)
∠HKF = ∠ABF = 64° (Corresponding angles)
∠FBC
= 180° - 64°
= 116° (Angles on a straight line)
∠BCL
= (180° - 116°) ÷ 2
= 32° (Isosceles triangle)
∠DCF
= 180° - 32°
= 148° (Angles on a straight line)
Answer(s): (a) 52°; (b) 148°