The figure is not drawn to scale. DFGH is a parallelogram, CDH and DEH are isosceles triangles. HG//KS and ∠GRS= 118°. CF is straight line.
- Find ∠CHD.
- Find ∠FEH.
(a)
∠GRS = 118°
∠HNP = 118° (Corresponding angles)
∠LNH
= 180° - 118°
= 62° (Angles on a straight line)
∠CHD
= 180° - 62° - 62°
= 56° (Isosceles triangle)
(b)
∠LNH = ∠CDH = 62° (Corresponding angles)
∠HDE
= 180° - 62°
= 118° (Angles on a straight line)
∠DEP
= (180° - 118°) ÷ 2
= 31° (Isosceles triangle)
∠FEH
= 180° - 31°
= 149° (Angles on a straight line)
Answer(s): (a) 56°; (b) 149°