The figure is not drawn to scale. CEFG is a parallelogram, BCG and CDG are isosceles triangles. GF//HR and ∠FPR= 104°. BE is straight line.
- Find ∠BGC.
- Find ∠EDG.
(a)
∠FPR = 104°
∠GLN = 104° (Corresponding angles)
∠KLG
= 180° - 104°
= 76° (Angles on a straight line)
∠BGC
= 180° - 76° - 76°
= 28° (Isosceles triangle)
(b)
∠KLG = ∠BCG = 76° (Corresponding angles)
∠GCD
= 180° - 76°
= 104° (Angles on a straight line)
∠CDN
= (180° - 104°) ÷ 2
= 38° (Isosceles triangle)
∠EDG
= 180° - 38°
= 142° (Angles on a straight line)
Answer(s): (a) 28°; (b) 142°