In the figure, NP = NR, ∠PNR = 33° and ∠SQR = 12°. Find
- ∠a
- ∠b
(a)
∠NPR
= (180° - 33°) ÷ 2
= 73.5° (Isosceles triangle)
∠a
= 180° - 12° - 73.5°
= 94.5° (Angles sum of triangle)
(b)
∠NRQ
= 180° - 73.5°
= 106.5° (Angles on a straight line)
∠b
= 12° + 106.5°
= 118.5° (Exterior angle of a triangle)
Answer(s): (a) 94.5°; (b) 118.5°