The figure is not drawn to scale. DFGH is a parallelogram, CDH and DEH are isosceles triangles. HG//KS and ∠GRS= 107°. CF is straight line.
- Find ∠CHD.
- Find ∠FEH.
(a)
∠GRS = 107°
∠HNP = 107° (Corresponding angles)
∠LNH
= 180° - 107°
= 73° (Angles on a straight line)
∠CHD
= 180° - 73° - 73°
= 34° (Isosceles triangle)
(b)
∠LNH = ∠CDH = 73° (Corresponding angles)
∠HDE
= 180° - 73°
= 107° (Angles on a straight line)
∠DEP
= (180° - 107°) ÷ 2
= 36.5° (Isosceles triangle)
∠FEH
= 180° - 36.5°
= 143.5° (Angles on a straight line)
Answer(s): (a) 34°; (b) 143.5°