The figure is not drawn to scale. CEFG is a parallelogram, BCG and CDG are isosceles triangles. GF//HR and ∠FPR= 111°. BE is straight line.
- Find ∠BGC.
- Find ∠EDG.
(a)
∠FPR = 111°
∠GLN = 111° (Corresponding angles)
∠KLG
= 180° - 111°
= 69° (Angles on a straight line)
∠BGC
= 180° - 69° - 69°
= 42° (Isosceles triangle)
(b)
∠KLG = ∠BCG = 69° (Corresponding angles)
∠GCD
= 180° - 69°
= 111° (Angles on a straight line)
∠CDN
= (180° - 111°) ÷ 2
= 34.5° (Isosceles triangle)
∠EDG
= 180° - 34.5°
= 145.5° (Angles on a straight line)
Answer(s): (a) 42°; (b) 145.5°