PSLE Lynn has a triangular piece of paper CDE with DC = DE, ∠CDE = 82° and ∠EFF = 66°. CFE and DFE are straight lines. She folded it along the line FG as shown.
- Find ∠h.
- Find ∠j.
(a)
Length of CD = Length of DE
Triangle CDE is an isosceles triangle.
∠DEC
= (180° - 82°) ÷ 2
= 98° ÷ 2
= 49° (Isosceles triangle)
∠h
= 180° - 66° - 49°
= 65° (Angles sum of triangle)
(b)
∠DCE = ∠DEC = 49°
∠k
= 180° - 66° - 66°
= 48° (Angles on a straight line)
∠j
= 180° - 49° - 48°
= 83° (Angles sum of triangle)
Answer(s): (a) 65°; (b) 83°