In the figure, EFGH is a parallelogram. GK, KL and EF are straight lines. EHJ is an isosceles triangle.
- Find ∠a.
- Find ∠b.
(a)
∠GKE = ∠FEL = 58° (Corresponding angles, EF//KG)
∠KJE
= 180° - ∠GKE - ∠KEJ
= 180° - 58° - 18°
= 104° (Angles sum of triangle)
∠HJE
= 180° - 104°
= 76° (Angles on a straight line)
∠JHE = ∠HJE = 76° (Isosceles triangle)
∠a
= ∠JHE
= 76° (Corresponding angles, FG//EH)
(b)
∠b
= 180° - 76° - 76°
= 28° (Isosceles triangle EHJ)
Answer(s): (a) 76°; (b) 28°
