Given that EFGH is a trapezium and EFH is an isosceles triangle, find the values of
- ∠f
- ∠g
(a)
∠f
= 180° - 85°
= 95° (Interior angles)
(b)
∠HFE
= 85° ÷ 2
= 42.5° (Exterior angle of a triangle)
∠GHF = ∠HFE = 42.5° (Alternate angles)
∠g
= 180° - 90° - 42.5°
= 47.5° (Angles sum of triangle)
Answer(s): (a) 95°; (b) 47.5°