The figure is not drawn to scale. BDEF is a parallelogram, ABF and BCF are isosceles triangles. FE//GP and ∠ENP= 117°. AD is straight line.
- Find ∠AFB.
- Find ∠DCF.
(a)
∠ENP = 117°
∠FKL = 117° (Corresponding angles)
∠HKF
= 180° - 117°
= 63° (Angles on a straight line)
∠AFB
= 180° - 63° - 63°
= 54° (Isosceles triangle)
(b)
∠HKF = ∠ABF = 63° (Corresponding angles)
∠FBC
= 180° - 63°
= 117° (Angles on a straight line)
∠BCL
= (180° - 117°) ÷ 2
= 31.5° (Isosceles triangle)
∠DCF
= 180° - 31.5°
= 148.5° (Angles on a straight line)
Answer(s): (a) 54°; (b) 148.5°