PSLE Pamela has a triangular piece of paper CDE with DC = DE, ∠CDE = 78° and ∠EFF = 74°. 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° - 78°) ÷ 2
= 102° ÷ 2
= 51° (Isosceles triangle)
∠h
= 180° - 74° - 51°
= 55° (Angles sum of triangle)
(b)
∠DCE = ∠DEC = 51°
∠k
= 180° - 74° - 74°
= 32° (Angles on a straight line)
∠j
= 180° - 51° - 32°
= 97° (Angles sum of triangle)
Answer(s): (a) 55°; (b) 97°