PSLE Yoko has a triangular piece of paper BCD with CB = CD, ∠BCD = 78° and ∠DEE = 74°. BED and CED are straight lines. She folded it along the line EF as shown.
- Find ∠g.
- Find ∠h.
(a)
Length of BC = Length of CD
Triangle BCD is an isosceles triangle.
∠CDB
= (180° - 78°) ÷ 2
= 102° ÷ 2
= 51° (Isosceles triangle)
∠g
= 180° - 74° - 51°
= 55° (Angles sum of triangle)
(b)
∠CBD = ∠CDB = 51°
∠j
= 180° - 74° - 74°
= 32° (Angles on a straight line)
∠h
= 180° - 51° - 32°
= 97° (Angles sum of triangle)
Answer(s): (a) 55°; (b) 97°