In the figure, NP = NR, ∠PNR = 35° and ∠SQR = 12°. Find
- ∠y
- ∠z
(a)
∠NPR
= (180° - 35°) ÷ 2
= 72.5° (Isosceles triangle)
∠y
= 180° - 12° - 72.5°
= 95.5° (Angles sum of triangle)
(b)
∠NRQ
= 180° - 72.5°
= 107.5° (Angles on a straight line)
∠z
= 12° + 107.5°
= 119.5° (Exterior angle of a triangle)
Answer(s): (a) 95.5°; (b) 119.5°