The figure is not drawn to scale. CEFG is a parallelogram, BCG and CDG are isosceles triangles. GF//HR and ∠FPR= 120°. BE is straight line.
- Find ∠BGC.
- Find ∠EDG.
(a)
∠FPR = 120°
∠GLN = 120° (Corresponding angles)
∠KLG
= 180° - 120°
= 60° (Angles on a straight line)
∠BGC
= 180° - 60° - 60°
= 60° (Isosceles triangle)
(b)
∠KLG = ∠BCG = 60° (Corresponding angles)
∠GCD
= 180° - 60°
= 120° (Angles on a straight line)
∠CDN
= (180° - 120°) ÷ 2
= 30° (Isosceles triangle)
∠EDG
= 180° - 30°
= 150° (Angles on a straight line)
Answer(s): (a) 60°; (b) 150°