In the figure, NP = NR, ∠PNR = 36° and ∠SQR = 15°. Find
- ∠p
- ∠q
(a)
∠NPR
= (180° - 36°) ÷ 2
= 72° (Isosceles triangle)
∠p
= 180° - 15° - 72°
= 93° (Angles sum of triangle)
(b)
∠NRQ
= 180° - 72°
= 108° (Angles on a straight line)
∠q
= 15° + 108°
= 123° (Exterior angle of a triangle)
Answer(s): (a) 93°; (b) 123°