The figure is not drawn to scale. DFGH is a parallelogram, CDH and DEH are isosceles triangles. HG//KS and ∠GRS= 101°. CF is straight line.
- Find ∠CHD.
- Find ∠FEH.
(a)
∠GRS = 101°
∠HNP = 101° (Corresponding angles)
∠LNH
= 180° - 101°
= 79° (Angles on a straight line)
∠CHD
= 180° - 79° - 79°
= 22° (Isosceles triangle)
(b)
∠LNH = ∠CDH = 79° (Corresponding angles)
∠HDE
= 180° - 79°
= 101° (Angles on a straight line)
∠DEP
= (180° - 101°) ÷ 2
= 39.5° (Isosceles triangle)
∠FEH
= 180° - 39.5°
= 140.5° (Angles on a straight line)
Answer(s): (a) 22°; (b) 140.5°