The figure is not drawn to scale. CEFG is a parallelogram, BCG and CDG are isosceles triangles. GF//HR and ∠FPR= 113°. BE is straight line.
- Find ∠BGC.
- Find ∠EDG.
(a)
∠FPR = 113°
∠GLN = 113° (Corresponding angles)
∠KLG
= 180° - 113°
= 67° (Angles on a straight line)
∠BGC
= 180° - 67° - 67°
= 46° (Isosceles triangle)
(b)
∠KLG = ∠BCG = 67° (Corresponding angles)
∠GCD
= 180° - 67°
= 113° (Angles on a straight line)
∠CDN
= (180° - 113°) ÷ 2
= 33.5° (Isosceles triangle)
∠EDG
= 180° - 33.5°
= 146.5° (Angles on a straight line)
Answer(s): (a) 46°; (b) 146.5°