DEFG is a rhombus, ABG and EBF are straight lines. If EB = BG, ∠BGF = 15° and ∠CEF = 33°, find
- ∠BFG
- ∠CAB
(a)
∠BFG
= 86° - 15°
= 71° (Exterior angle of a triangle)
(b)
∠EAB
= 86° - 33°
= 53° (Exterior angle of a triangle)
∠CAB
= 180° - 53°
= 127° (Angles on a straight line)
Answer(s): (a) 71°; (b) 127°