The figure is not drawn to scale. DFGH is a parallelogram, CDH and DEH are isosceles triangles. HG//KS and ∠GRS= 112°. CF is straight line.
- Find ∠CHD.
- Find ∠FEH.
(a)
∠GRS = 112°
∠HNP = 112° (Corresponding angles)
∠LNH
= 180° - 112°
= 68° (Angles on a straight line)
∠CHD
= 180° - 68° - 68°
= 44° (Isosceles triangle)
(b)
∠LNH = ∠CDH = 68° (Corresponding angles)
∠HDE
= 180° - 68°
= 112° (Angles on a straight line)
∠DEP
= (180° - 112°) ÷ 2
= 34° (Isosceles triangle)
∠FEH
= 180° - 34°
= 146° (Angles on a straight line)
Answer(s): (a) 44°; (b) 146°
