DEFG is a rhombus, ABG and EBF are straight lines. If EB = BG, ∠BGF = 21° and ∠CEF = 34°, find
- ∠BFG
- ∠CAB
(a)
∠BFG
= 87° - 21°
= 66° (Exterior angle of a triangle)
(b)
∠EAB
= 87° - 34°
= 53° (Exterior angle of a triangle)
∠CAB
= 180° - 53°
= 127° (Angles on a straight line)
Answer(s): (a) 66°; (b) 127°