In the figure, EF = EH, ∠FEH = 31° and ∠JGH = 13°. Find
- ∠p
- ∠q
(a)
∠EFH
= (180° - 31°) ÷ 2
= 74.5° (Isosceles triangle)
∠p
= 180° - 13° - 74.5°
= 92.5° (Angles sum of triangle)
(b)
∠EHG
= 180° - 74.5°
= 105.5° (Angles on a straight line)
∠q
= 13° + 105.5°
= 118.5° (Exterior angle of a triangle)
Answer(s): (a) 92.5°; (b) 118.5°