PSLE Tammy has a triangular piece of paper EFG with FE = FG, ∠EFG = 76° and ∠GHH = 73°. 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° - 76°) ÷ 2
= 104° ÷ 2
= 52° (Isosceles triangle)
∠k
= 180° - 73° - 52°
= 55° (Angles sum of triangle)
(b)
∠FEG = ∠FGE = 52°
∠m
= 180° - 73° - 73°
= 34° (Angles on a straight line)
∠l
= 180° - 52° - 34°
= 94° (Angles sum of triangle)
Answer(s): (a) 55°; (b) 94°