The figure is not drawn to scale. CEFG is a parallelogram, BCG and CDG are isosceles triangles. GF//HR and ∠FPR= 101°. BE is straight line.
- Find ∠BGC.
- Find ∠EDG.
(a)
∠FPR = 101°
∠GLN = 101° (Corresponding angles)
∠KLG
= 180° - 101°
= 79° (Angles on a straight line)
∠BGC
= 180° - 79° - 79°
= 22° (Isosceles triangle)
(b)
∠KLG = ∠BCG = 79° (Corresponding angles)
∠GCD
= 180° - 79°
= 101° (Angles on a straight line)
∠CDN
= (180° - 101°) ÷ 2
= 39.5° (Isosceles triangle)
∠EDG
= 180° - 39.5°
= 140.5° (Angles on a straight line)
Answer(s): (a) 22°; (b) 140.5°