The figure is not drawn to scale. DFGH is a parallelogram, CDH and DEH are isosceles triangles. HG//KS and ∠GRS= 117°. CF is straight line.
- Find ∠CHD.
- Find ∠FEH.
(a)
∠GRS = 117°
∠HNP = 117° (Corresponding angles)
∠LNH
= 180° - 117°
= 63° (Angles on a straight line)
∠CHD
= 180° - 63° - 63°
= 54° (Isosceles triangle)
(b)
∠LNH = ∠CDH = 63° (Corresponding angles)
∠HDE
= 180° - 63°
= 117° (Angles on a straight line)
∠DEP
= (180° - 117°) ÷ 2
= 31.5° (Isosceles triangle)
∠FEH
= 180° - 31.5°
= 148.5° (Angles on a straight line)
Answer(s): (a) 54°; (b) 148.5°