The figure is not drawn to scale. DFGH is a parallelogram, CDH and DEH are isosceles triangles. HG//KS and ∠GRS= 102°. CF is straight line.
- Find ∠CHD.
- Find ∠FEH.
(a)
∠GRS = 102°
∠HNP = 102° (Corresponding angles)
∠LNH
= 180° - 102°
= 78° (Angles on a straight line)
∠CHD
= 180° - 78° - 78°
= 24° (Isosceles triangle)
(b)
∠LNH = ∠CDH = 78° (Corresponding angles)
∠HDE
= 180° - 78°
= 102° (Angles on a straight line)
∠DEP
= (180° - 102°) ÷ 2
= 39° (Isosceles triangle)
∠FEH
= 180° - 39°
= 141° (Angles on a straight line)
Answer(s): (a) 24°; (b) 141°