DEFG is a rhombus, ABG and EBF are straight lines. If EB = BG, ∠BGF = 15° and ∠CEF = 33°, find
- ∠BFG
- ∠CAB
(a)
∠BFG
= 88° - 15°
= 73° (Exterior angle of a triangle)
(b)
∠EAB
= 88° - 33°
= 55° (Exterior angle of a triangle)
∠CAB
= 180° - 55°
= 125° (Angles on a straight line)
Answer(s): (a) 73°; (b) 125°