PSLE Barbara has a triangular piece of paper JKL with KJ = KL, ∠JKL = 80° and ∠LMM = 72°. JML and KML are straight lines. She folded it along the line MN as shown.
- Find ∠p.
- Find ∠q.
(a)
Length of JK = Length of KL
Triangle JKL is an isosceles triangle.
∠KLJ
= (180° - 80°) ÷ 2
= 100° ÷ 2
= 50° (Isosceles triangle)
∠p
= 180° - 72° - 50°
= 58° (Angles sum of triangle)
(b)
∠KJL = ∠KLJ = 50°
∠r
= 180° - 72° - 72°
= 36° (Angles on a straight line)
∠q
= 180° - 50° - 36°
= 94° (Angles sum of triangle)
Answer(s): (a) 58°; (b) 94°