PSLE Min has a triangular piece of paper CDE with DC = DE, ∠CDE = 80° 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° - 80°) ÷ 2
= 100° ÷ 2
= 50° (Isosceles triangle)
∠h
= 180° - 74° - 50°
= 56° (Angles sum of triangle)
(b)
∠DCE = ∠DEC = 50°
∠k
= 180° - 74° - 74°
= 32° (Angles on a straight line)
∠j
= 180° - 50° - 32°
= 98° (Angles sum of triangle)
Answer(s): (a) 56°; (b) 98°