In the figure, EFGH is a parallelogram. GK, KL and EF are straight lines. EHJ is an isosceles triangle.
- Find ∠g.
- Find ∠h.
(a)
∠GKE = ∠FEL = 60° (Corresponding angles, EF//KG)
∠KJE
= 180° - ∠GKE - ∠KEJ
= 180° - 60° - 21°
= 99° (Angles sum of triangle)
∠HJE
= 180° - 99°
= 81° (Angles on a straight line)
∠JHE = ∠HJE = 81° (Isosceles triangle)
∠g
= ∠JHE
= 81° (Corresponding angles, FG//EH)
(b)
∠h
= 180° - 81° - 81°
= 18° (Isosceles triangle EHJ)
Answer(s): (a) 81°; (b) 18°