In the figure, NP = NR, ∠PNR = 35° and ∠SQR = 17°. Find
- ∠q
- ∠r
(a)
∠NPR
= (180° - 35°) ÷ 2
= 72.5° (Isosceles triangle)
∠q
= 180° - 17° - 72.5°
= 90.5° (Angles sum of triangle)
(b)
∠NRQ
= 180° - 72.5°
= 107.5° (Angles on a straight line)
∠r
= 17° + 107.5°
= 124.5° (Exterior angle of a triangle)
Answer(s): (a) 90.5°; (b) 124.5°