The figure is not drawn to scale. DFGH is a parallelogram, CDH and DEH are isosceles triangles. HG//KS and ∠GRS= 104°. CF is straight line.
- Find ∠CHD.
- Find ∠FEH.
(a)
∠GRS = 104°
∠HNP = 104° (Corresponding angles)
∠LNH
= 180° - 104°
= 76° (Angles on a straight line)
∠CHD
= 180° - 76° - 76°
= 28° (Isosceles triangle)
(b)
∠LNH = ∠CDH = 76° (Corresponding angles)
∠HDE
= 180° - 76°
= 104° (Angles on a straight line)
∠DEP
= (180° - 104°) ÷ 2
= 38° (Isosceles triangle)
∠FEH
= 180° - 38°
= 142° (Angles on a straight line)
Answer(s): (a) 28°; (b) 142°