In the figure, NP = NR, ∠PNR = 32° and ∠SQR = 15°. Find
- ∠b
- ∠c
(a)
∠NPR
= (180° - 32°) ÷ 2
= 74° (Isosceles triangle)
∠b
= 180° - 15° - 74°
= 91° (Angles sum of triangle)
(b)
∠NRQ
= 180° - 74°
= 106° (Angles on a straight line)
∠c
= 15° + 106°
= 121° (Exterior angle of a triangle)
Answer(s): (a) 91°; (b) 121°