The figure is not drawn to scale. CEFG is a parallelogram, BCG and CDG are isosceles triangles. GF//HR and ∠FPR= 108°. BE is straight line.
- Find ∠BGC.
- Find ∠EDG.
(a)
∠FPR = 108°
∠GLN = 108° (Corresponding angles)
∠KLG
= 180° - 108°
= 72° (Angles on a straight line)
∠BGC
= 180° - 72° - 72°
= 36° (Isosceles triangle)
(b)
∠KLG = ∠BCG = 72° (Corresponding angles)
∠GCD
= 180° - 72°
= 108° (Angles on a straight line)
∠CDN
= (180° - 108°) ÷ 2
= 36° (Isosceles triangle)
∠EDG
= 180° - 36°
= 144° (Angles on a straight line)
Answer(s): (a) 36°; (b) 144°