PSLE Marion has a triangular piece of paper NPQ with PN = PQ, ∠NPQ = 76° and ∠QRR = 72°. NRQ and PRQ are straight lines. She folded it along the line RS as shown.
- Find ∠t.
- Find ∠u.
(a)
Length of NP = Length of PQ
Triangle NPQ is an isosceles triangle.
∠PQN
= (180° - 76°) ÷ 2
= 104° ÷ 2
= 52° (Isosceles triangle)
∠t
= 180° - 72° - 52°
= 56° (Angles sum of triangle)
(b)
∠PNQ = ∠PQN = 52°
∠v
= 180° - 72° - 72°
= 36° (Angles on a straight line)
∠u
= 180° - 52° - 36°
= 92° (Angles sum of triangle)
Answer(s): (a) 56°; (b) 92°