PSLE Usha has a triangular piece of paper NPQ with PN = PQ, ∠NPQ = 80° and ∠QRR = 74°. 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° - 80°) ÷ 2
= 100° ÷ 2
= 50° (Isosceles triangle)
∠t
= 180° - 74° - 50°
= 56° (Angles sum of triangle)
(b)
∠PNQ = ∠PQN = 50°
∠v
= 180° - 74° - 74°
= 32° (Angles on a straight line)
∠u
= 180° - 50° - 32°
= 98° (Angles sum of triangle)
Answer(s): (a) 56°; (b) 98°