Given that BCDE is a rhombus, calculate
- ∠h
- ∠k
(a)
∠BFE = ∠CFG = 104° (Vertically opposite angles)
∠h
= 180° - 104° - 16°
= 60° (Angles sum of triangle)
(b)
∠EDC
= 180° - 16° - 16°
= 148° (Angles sum of triangle)
∠k
= 360° - 104° - 148° - 16°
= 92° (Sum of Angles in a quadrilateral)
Answer(s): (a) 60°; (b) 92°