EFGH is a rhombus, BCH and FCG are straight lines. If FC = CH, ∠CHG = 15° and ∠DFG = 35°, find
- ∠CGH
- ∠DBC
(a)
∠CGH
= 85° - 15°
= 70° (Exterior angle of a triangle)
(b)
∠FBC
= 85° - 35°
= 50° (Exterior angle of a triangle)
∠DBC
= 180° - 50°
= 130° (Angles on a straight line)
Answer(s): (a) 70°; (b) 130°