The figure is not drawn to scale. CEFG is a parallelogram, BCG and CDG are isosceles triangles. GF//HR and ∠FPR= 118°. BE is straight line.
- Find ∠BGC.
- Find ∠EDG.
(a)
∠FPR = 118°
∠GLN = 118° (Corresponding angles)
∠KLG
= 180° - 118°
= 62° (Angles on a straight line)
∠BGC
= 180° - 62° - 62°
= 56° (Isosceles triangle)
(b)
∠KLG = ∠BCG = 62° (Corresponding angles)
∠GCD
= 180° - 62°
= 118° (Angles on a straight line)
∠CDN
= (180° - 118°) ÷ 2
= 31° (Isosceles triangle)
∠EDG
= 180° - 31°
= 149° (Angles on a straight line)
Answer(s): (a) 56°; (b) 149°
