The figure is not drawn to scale. BDEF is a parallelogram, ABF and BCF are isosceles triangles. FE//GP and ∠ENP= 118°. AD is straight line.
- Find ∠AFB.
- Find ∠DCF.
(a)
∠ENP = 118°
∠FKL = 118° (Corresponding angles)
∠HKF
= 180° - 118°
= 62° (Angles on a straight line)
∠AFB
= 180° - 62° - 62°
= 56° (Isosceles triangle)
(b)
∠HKF = ∠ABF = 62° (Corresponding angles)
∠FBC
= 180° - 62°
= 118° (Angles on a straight line)
∠BCL
= (180° - 118°) ÷ 2
= 31° (Isosceles triangle)
∠DCF
= 180° - 31°
= 149° (Angles on a straight line)
Answer(s): (a) 56°; (b) 149°