The figure is not drawn to scale. DFGH is a parallelogram, CDH and DEH are isosceles triangles. HG//KS and ∠GRS= 116°. CF is straight line.
- Find ∠CHD.
- Find ∠FEH.
(a)
∠GRS = 116°
∠HNP = 116° (Corresponding angles)
∠LNH
= 180° - 116°
= 64° (Angles on a straight line)
∠CHD
= 180° - 64° - 64°
= 52° (Isosceles triangle)
(b)
∠LNH = ∠CDH = 64° (Corresponding angles)
∠HDE
= 180° - 64°
= 116° (Angles on a straight line)
∠DEP
= (180° - 116°) ÷ 2
= 32° (Isosceles triangle)
∠FEH
= 180° - 32°
= 148° (Angles on a straight line)
Answer(s): (a) 52°; (b) 148°