Given that EFGH is a rhombus, calculate
- ∠n
- ∠p
(a)
∠EKH = ∠FKL = 98° (Vertically opposite angles)
∠n
= 180° - 98° - 25°
= 57° (Angles sum of triangle)
(b)
∠HGF
= 180° - 25° - 25°
= 130° (Angles sum of triangle)
∠p
= 360° - 98° - 130° - 25°
= 107° (Sum of Angles in a quadrilateral)
Answer(s): (a) 57°; (b) 107°