In the figure, EF = EH, ∠FEH = 32° and ∠JGH = 18°. Find
- ∠h
- ∠i
(a)
∠EFH
= (180° - 32°) ÷ 2
= 74° (Isosceles triangle)
∠h
= 180° - 18° - 74°
= 88° (Angles sum of triangle)
(b)
∠EHG
= 180° - 74°
= 106° (Angles on a straight line)
∠i
= 18° + 106°
= 124° (Exterior angle of a triangle)
Answer(s): (a) 88°; (b) 124°