PSLE Shannon has a triangular piece of paper CDE with DC = DE, ∠CDE = 76° and ∠EFF = 73°. 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° - 76°) ÷ 2
= 104° ÷ 2
= 52° (Isosceles triangle)
∠h
= 180° - 73° - 52°
= 55° (Angles sum of triangle)
(b)
∠DCE = ∠DEC = 52°
∠k
= 180° - 73° - 73°
= 34° (Angles on a straight line)
∠j
= 180° - 52° - 34°
= 94° (Angles sum of triangle)
Answer(s): (a) 55°; (b) 94°