Given that EFGH is a trapezium and EFH is an isosceles triangle, find the values of
- ∠n
- ∠p
(a)
∠n
= 180° - 83°
= 97° (Interior angles)
(b)
∠HFE
= 83° ÷ 2
= 41.5° (Exterior angle of a triangle)
∠GHF = ∠HFE = 41.5° (Alternate angles)
∠p
= 180° - 90° - 41.5°
= 48.5° (Angles sum of triangle)
Answer(s): (a) 97°; (b) 48.5°