The figure is not drawn to scale. CEFG is a parallelogram, BCG and CDG are isosceles triangles. GF//HR and ∠FPR= 115°. BE is straight line.
- Find ∠BGC.
- Find ∠EDG.
(a)
∠FPR = 115°
∠GLN = 115° (Corresponding angles)
∠KLG
= 180° - 115°
= 65° (Angles on a straight line)
∠BGC
= 180° - 65° - 65°
= 50° (Isosceles triangle)
(b)
∠KLG = ∠BCG = 65° (Corresponding angles)
∠GCD
= 180° - 65°
= 115° (Angles on a straight line)
∠CDN
= (180° - 115°) ÷ 2
= 32.5° (Isosceles triangle)
∠EDG
= 180° - 32.5°
= 147.5° (Angles on a straight line)
Answer(s): (a) 50°; (b) 147.5°