The figure is not drawn to scale. DFGH is a parallelogram, CDH and DEH are isosceles triangles. HG//KS and ∠GRS= 100°. CF is straight line.
- Find ∠CHD.
- Find ∠FEH.
(a)
∠GRS = 100°
∠HNP = 100° (Corresponding angles)
∠LNH
= 180° - 100°
= 80° (Angles on a straight line)
∠CHD
= 180° - 80° - 80°
= 20° (Isosceles triangle)
(b)
∠LNH = ∠CDH = 80° (Corresponding angles)
∠HDE
= 180° - 80°
= 100° (Angles on a straight line)
∠DEP
= (180° - 100°) ÷ 2
= 40° (Isosceles triangle)
∠FEH
= 180° - 40°
= 140° (Angles on a straight line)
Answer(s): (a) 20°; (b) 140°