In the figure, NP = NR, ∠PNR = 35° and ∠SQR = 13°. Find
- ∠a
- ∠b
(a)
∠NPR
= (180° - 35°) ÷ 2
= 72.5° (Isosceles triangle)
∠a
= 180° - 13° - 72.5°
= 94.5° (Angles sum of triangle)
(b)
∠NRQ
= 180° - 72.5°
= 107.5° (Angles on a straight line)
∠b
= 13° + 107.5°
= 120.5° (Exterior angle of a triangle)
Answer(s): (a) 94.5°; (b) 120.5°