The figure is not drawn to scale. DFGH is a parallelogram, CDH and DEH are isosceles triangles. HG//KS and ∠GRS= 103°. CF is straight line.
- Find ∠CHD.
- Find ∠FEH.
(a)
∠GRS = 103°
∠HNP = 103° (Corresponding angles)
∠LNH
= 180° - 103°
= 77° (Angles on a straight line)
∠CHD
= 180° - 77° - 77°
= 26° (Isosceles triangle)
(b)
∠LNH = ∠CDH = 77° (Corresponding angles)
∠HDE
= 180° - 77°
= 103° (Angles on a straight line)
∠DEP
= (180° - 103°) ÷ 2
= 38.5° (Isosceles triangle)
∠FEH
= 180° - 38.5°
= 141.5° (Angles on a straight line)
Answer(s): (a) 26°; (b) 141.5°