PSLE Usha has a triangular piece of paper KLM with LK = LM, ∠KLM = 82° and ∠MNN = 75°. KNM and LNM are straight lines. She folded it along the line NP as shown.
- Find ∠q.
- Find ∠r.
(a)
Length of KL = Length of LM
Triangle KLM is an isosceles triangle.
∠LMK
= (180° - 82°) ÷ 2
= 98° ÷ 2
= 49° (Isosceles triangle)
∠q
= 180° - 75° - 49°
= 56° (Angles sum of triangle)
(b)
∠LKM = ∠LMK = 49°
∠s
= 180° - 75° - 75°
= 30° (Angles on a straight line)
∠r
= 180° - 49° - 30°
= 101° (Angles sum of triangle)
Answer(s): (a) 56°; (b) 101°