The figure is not drawn to scale. CEFG is a parallelogram, BCG and CDG are isosceles triangles. GF//HR and ∠FPR= 103°. BE is straight line.
- Find ∠BGC.
- Find ∠EDG.
(a)
∠FPR = 103°
∠GLN = 103° (Corresponding angles)
∠KLG
= 180° - 103°
= 77° (Angles on a straight line)
∠BGC
= 180° - 77° - 77°
= 26° (Isosceles triangle)
(b)
∠KLG = ∠BCG = 77° (Corresponding angles)
∠GCD
= 180° - 77°
= 103° (Angles on a straight line)
∠CDN
= (180° - 103°) ÷ 2
= 38.5° (Isosceles triangle)
∠EDG
= 180° - 38.5°
= 141.5° (Angles on a straight line)
Answer(s): (a) 26°; (b) 141.5°