The figure is not drawn to scale. DEGH is a parallelogram. ∠HDO is 56° and ∠GEF is 29°. DEF is an isosceles triangle and DE = EF.
- Find ∠OFE.
- Find ∠DHG.
(a)
∠EOD = 56° (Alternate angles, DH//EG)
∠EOF
= 180° - 56°
= 124° (Angles on a straight line)
∠OFE
= 180° - 124° - 29°
= 27° (Angles sum of triangle)
(b)
∠EDO = 27° (Isosceles triangle DEF)
∠DHG
= 180° - 56° - 27°
= 97° (Interior angles, DE//HG)
Answer(s): (a) 27°; (b) 97°