PSLE Erika has a triangular piece of paper EFG with FE = FG, ∠EFG = 80° and ∠GHH = 71°. EHG and FHG are straight lines. She folded it along the line HJ as shown.
- Find ∠k.
- Find ∠l.
(a)
Length of EF = Length of FG
Triangle EFG is an isosceles triangle.
∠FGE
= (180° - 80°) ÷ 2
= 100° ÷ 2
= 50° (Isosceles triangle)
∠k
= 180° - 71° - 50°
= 59° (Angles sum of triangle)
(b)
∠FEG = ∠FGE = 50°
∠m
= 180° - 71° - 71°
= 38° (Angles on a straight line)
∠l
= 180° - 50° - 38°
= 92° (Angles sum of triangle)
Answer(s): (a) 59°; (b) 92°