The figure is not drawn to scale. BDEF is a parallelogram, ABF and BCF are isosceles triangles. FE//GP and ∠ENP= 102°. AD is straight line.
- Find ∠AFB.
- Find ∠DCF.
(a)
∠ENP = 102°
∠FKL = 102° (Corresponding angles)
∠HKF
= 180° - 102°
= 78° (Angles on a straight line)
∠AFB
= 180° - 78° - 78°
= 24° (Isosceles triangle)
(b)
∠HKF = ∠ABF = 78° (Corresponding angles)
∠FBC
= 180° - 78°
= 102° (Angles on a straight line)
∠BCL
= (180° - 102°) ÷ 2
= 39° (Isosceles triangle)
∠DCF
= 180° - 39°
= 141° (Angles on a straight line)
Answer(s): (a) 24°; (b) 141°