PSLE Rachel has a triangular piece of paper CDE with DC = DE, ∠CDE = 84° and ∠EFF = 69°. 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° - 84°) ÷ 2
= 96° ÷ 2
= 48° (Isosceles triangle)
∠h
= 180° - 69° - 48°
= 63° (Angles sum of triangle)
(b)
∠DCE = ∠DEC = 48°
∠k
= 180° - 69° - 69°
= 42° (Angles on a straight line)
∠j
= 180° - 48° - 42°
= 90° (Angles sum of triangle)
Answer(s): (a) 63°; (b) 90°