PSLE Dana has a triangular piece of paper JKL with KJ = KL, ∠JKL = 78° and ∠LMM = 65°. 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° - 78°) ÷ 2
= 102° ÷ 2
= 51° (Isosceles triangle)
∠p
= 180° - 65° - 51°
= 64° (Angles sum of triangle)
(b)
∠KJL = ∠KLJ = 51°
∠r
= 180° - 65° - 65°
= 50° (Angles on a straight line)
∠q
= 180° - 51° - 50°
= 79° (Angles sum of triangle)
Answer(s): (a) 64°; (b) 79°