DEFG is a rhombus, ABG and EBF are straight lines. If EB = BG, ∠BGF = 21° and ∠CEF = 32°, find
- ∠BFG
- ∠CAB
(a)
∠BFG
= 86° - 21°
= 65° (Exterior angle of a triangle)
(b)
∠EAB
= 86° - 32°
= 54° (Exterior angle of a triangle)
∠CAB
= 180° - 54°
= 126° (Angles on a straight line)
Answer(s): (a) 65°; (b) 126°