The figure is not drawn to scale. CEFG is a parallelogram, BCG and CDG are isosceles triangles. GF//HR and ∠FPR= 107°. BE is straight line.
- Find ∠BGC.
- Find ∠EDG.
(a)
∠FPR = 107°
∠GLN = 107° (Corresponding angles)
∠KLG
= 180° - 107°
= 73° (Angles on a straight line)
∠BGC
= 180° - 73° - 73°
= 34° (Isosceles triangle)
(b)
∠KLG = ∠BCG = 73° (Corresponding angles)
∠GCD
= 180° - 73°
= 107° (Angles on a straight line)
∠CDN
= (180° - 107°) ÷ 2
= 36.5° (Isosceles triangle)
∠EDG
= 180° - 36.5°
= 143.5° (Angles on a straight line)
Answer(s): (a) 34°; (b) 143.5°