The figure is not drawn to scale. CEFG is a parallelogram, BCG and CDG are isosceles triangles. GF//HR and ∠FPR= 100°. BE is straight line.
- Find ∠BGC.
- Find ∠EDG.
(a)
∠FPR = 100°
∠GLN = 100° (Corresponding angles)
∠KLG
= 180° - 100°
= 80° (Angles on a straight line)
∠BGC
= 180° - 80° - 80°
= 20° (Isosceles triangle)
(b)
∠KLG = ∠BCG = 80° (Corresponding angles)
∠GCD
= 180° - 80°
= 100° (Angles on a straight line)
∠CDN
= (180° - 100°) ÷ 2
= 40° (Isosceles triangle)
∠EDG
= 180° - 40°
= 140° (Angles on a straight line)
Answer(s): (a) 20°; (b) 140°