In the figure, EF = EH, ∠FEH = 32° and ∠JGH = 16°. Find
- ∠h
- ∠i
(a)
∠EFH
= (180° - 32°) ÷ 2
= 74° (Isosceles triangle)
∠h
= 180° - 16° - 74°
= 90° (Angles sum of triangle)
(b)
∠EHG
= 180° - 74°
= 106° (Angles on a straight line)
∠i
= 16° + 106°
= 122° (Exterior angle of a triangle)
Answer(s): (a) 90°; (b) 122°