The figure is not drawn to scale. CEFG is a parallelogram, BCG and CDG are isosceles triangles. GF//HR and ∠FPR= 119°. BE is straight line.
- Find ∠BGC.
- Find ∠EDG.
(a)
∠FPR = 119°
∠GLN = 119° (Corresponding angles)
∠KLG
= 180° - 119°
= 61° (Angles on a straight line)
∠BGC
= 180° - 61° - 61°
= 58° (Isosceles triangle)
(b)
∠KLG = ∠BCG = 61° (Corresponding angles)
∠GCD
= 180° - 61°
= 119° (Angles on a straight line)
∠CDN
= (180° - 119°) ÷ 2
= 30.5° (Isosceles triangle)
∠EDG
= 180° - 30.5°
= 149.5° (Angles on a straight line)
Answer(s): (a) 58°; (b) 149.5°