In the figure, EF = EH, ∠FEH = 33° and ∠JGH = 17°. Find
- ∠e
- ∠f
(a)
∠EFH
= (180° - 33°) ÷ 2
= 73.5° (Isosceles triangle)
∠e
= 180° - 17° - 73.5°
= 89.5° (Angles sum of triangle)
(b)
∠EHG
= 180° - 73.5°
= 106.5° (Angles on a straight line)
∠f
= 17° + 106.5°
= 123.5° (Exterior angle of a triangle)
Answer(s): (a) 89.5°; (b) 123.5°