The figure is not drawn to scale. CEFG is a parallelogram, BCG and CDG are isosceles triangles. GF//HR and ∠FPR= 112°. BE is straight line.
- Find ∠BGC.
- Find ∠EDG.
(a)
∠FPR = 112°
∠GLN = 112° (Corresponding angles)
∠KLG
= 180° - 112°
= 68° (Angles on a straight line)
∠BGC
= 180° - 68° - 68°
= 44° (Isosceles triangle)
(b)
∠KLG = ∠BCG = 68° (Corresponding angles)
∠GCD
= 180° - 68°
= 112° (Angles on a straight line)
∠CDN
= (180° - 112°) ÷ 2
= 34° (Isosceles triangle)
∠EDG
= 180° - 34°
= 146° (Angles on a straight line)
Answer(s): (a) 44°; (b) 146°