The figure is not drawn to scale. BDEF is a parallelogram, ABF and BCF are isosceles triangles. FE//GP and ∠ENP= 107°. AD is straight line.
- Find ∠AFB.
- Find ∠DCF.
(a)
∠ENP = 107°
∠FKL = 107° (Corresponding angles)
∠HKF
= 180° - 107°
= 73° (Angles on a straight line)
∠AFB
= 180° - 73° - 73°
= 34° (Isosceles triangle)
(b)
∠HKF = ∠ABF = 73° (Corresponding angles)
∠FBC
= 180° - 73°
= 107° (Angles on a straight line)
∠BCL
= (180° - 107°) ÷ 2
= 36.5° (Isosceles triangle)
∠DCF
= 180° - 36.5°
= 143.5° (Angles on a straight line)
Answer(s): (a) 34°; (b) 143.5°