In the figure, EF = EH, ∠FEH = 35° and ∠JGH = 18°. Find
- ∠q
- ∠r
(a)
∠EFH
= (180° - 35°) ÷ 2
= 72.5° (Isosceles triangle)
∠q
= 180° - 18° - 72.5°
= 89.5° (Angles sum of triangle)
(b)
∠EHG
= 180° - 72.5°
= 107.5° (Angles on a straight line)
∠r
= 18° + 107.5°
= 125.5° (Exterior angle of a triangle)
Answer(s): (a) 89.5°; (b) 125.5°