The figure is not drawn to scale. BDEF is a parallelogram, ABF and BCF are isosceles triangles. FE//GP and ∠ENP= 108°. AD is straight line.
- Find ∠AFB.
- Find ∠DCF.
(a)
∠ENP = 108°
∠FKL = 108° (Corresponding angles)
∠HKF
= 180° - 108°
= 72° (Angles on a straight line)
∠AFB
= 180° - 72° - 72°
= 36° (Isosceles triangle)
(b)
∠HKF = ∠ABF = 72° (Corresponding angles)
∠FBC
= 180° - 72°
= 108° (Angles on a straight line)
∠BCL
= (180° - 108°) ÷ 2
= 36° (Isosceles triangle)
∠DCF
= 180° - 36°
= 144° (Angles on a straight line)
Answer(s): (a) 36°; (b) 144°