The figure is not drawn to scale. EFHJ is a parallelogram. ∠JEO is 56° and ∠HFG is 32°. EFG is an isosceles triangle and EF = FG.
- Find ∠OGF.
- Find ∠EJH.
(a)
∠FOE = 56° (Alternate angles, EJ//FH)
∠FOG
= 180° - 56°
= 124° (Angles on a straight line)
∠OGF
= 180° - 124° - 32°
= 24° (Angles sum of triangle)
(b)
∠FEO = 24° (Isosceles triangle EFG)
∠EJH
= 180° - 56° - 24°
= 100° (Interior angles, EF//JH)
Answer(s): (a) 24°; (b) 100°