In the figure, EF = EH, ∠FEH = 36° and ∠JGH = 13°. Find
- ∠y
- ∠z
(a)
∠EFH
= (180° - 36°) ÷ 2
= 72° (Isosceles triangle)
∠y
= 180° - 13° - 72°
= 95° (Angles sum of triangle)
(b)
∠EHG
= 180° - 72°
= 108° (Angles on a straight line)
∠z
= 13° + 108°
= 121° (Exterior angle of a triangle)
Answer(s): (a) 95°; (b) 121°