The figure is not drawn to scale. DFGH is a parallelogram, CDH and DEH are isosceles triangles. HG//KS and ∠GRS= 109°. CF is straight line.
- Find ∠CHD.
- Find ∠FEH.
(a)
∠GRS = 109°
∠HNP = 109° (Corresponding angles)
∠LNH
= 180° - 109°
= 71° (Angles on a straight line)
∠CHD
= 180° - 71° - 71°
= 38° (Isosceles triangle)
(b)
∠LNH = ∠CDH = 71° (Corresponding angles)
∠HDE
= 180° - 71°
= 109° (Angles on a straight line)
∠DEP
= (180° - 109°) ÷ 2
= 35.5° (Isosceles triangle)
∠FEH
= 180° - 35.5°
= 144.5° (Angles on a straight line)
Answer(s): (a) 38°; (b) 144.5°