The figure is not drawn to scale. DFGH is a parallelogram, CDH and DEH are isosceles triangles. HG//KS and ∠GRS= 119°. CF is straight line.
- Find ∠CHD.
- Find ∠FEH.
(a)
∠GRS = 119°
∠HNP = 119° (Corresponding angles)
∠LNH
= 180° - 119°
= 61° (Angles on a straight line)
∠CHD
= 180° - 61° - 61°
= 58° (Isosceles triangle)
(b)
∠LNH = ∠CDH = 61° (Corresponding angles)
∠HDE
= 180° - 61°
= 119° (Angles on a straight line)
∠DEP
= (180° - 119°) ÷ 2
= 30.5° (Isosceles triangle)
∠FEH
= 180° - 30.5°
= 149.5° (Angles on a straight line)
Answer(s): (a) 58°; (b) 149.5°