Given that EFGH is a rhombus, calculate
- ∠n
- ∠p
(a)
∠EKH = ∠FKL = 104° (Vertically opposite angles)
∠n
= 180° - 104° - 12°
= 64° (Angles sum of triangle)
(b)
∠HGF
= 180° - 12° - 12°
= 156° (Angles sum of triangle)
∠p
= 360° - 104° - 156° - 12°
= 88° (Sum of Angles in a quadrilateral)
Answer(s): (a) 64°; (b) 88°