In the figure, NP = NR, ∠PNR = 35° and ∠SQR = 16°. Find
- ∠h
- ∠i
(a)
∠NPR
= (180° - 35°) ÷ 2
= 72.5° (Isosceles triangle)
∠h
= 180° - 16° - 72.5°
= 91.5° (Angles sum of triangle)
(b)
∠NRQ
= 180° - 72.5°
= 107.5° (Angles on a straight line)
∠i
= 16° + 107.5°
= 123.5° (Exterior angle of a triangle)
Answer(s): (a) 91.5°; (b) 123.5°