In the figure, EFGH is a parallelogram. GK, KL and EF are straight lines. EHJ is an isosceles triangle.
- Find ∠h.
- Find ∠i.
(a)
∠GKE = ∠FEL = 65° (Corresponding angles, EF//KG)
∠KJE
= 180° - ∠GKE - ∠KEJ
= 180° - 65° - 15°
= 100° (Angles sum of triangle)
∠HJE
= 180° - 100°
= 80° (Angles on a straight line)
∠JHE = ∠HJE = 80° (Isosceles triangle)
∠h
= ∠JHE
= 80° (Corresponding angles, FG//EH)
(b)
∠i
= 180° - 80° - 80°
= 20° (Isosceles triangle EHJ)
Answer(s): (a) 80°; (b) 20°