In the figure, NP = NR, ∠PNR = 34° and ∠SQR = 16°. Find
- ∠g
- ∠h
(a)
∠NPR
= (180° - 34°) ÷ 2
= 73° (Isosceles triangle)
∠g
= 180° - 16° - 73°
= 91° (Angles sum of triangle)
(b)
∠NRQ
= 180° - 73°
= 107° (Angles on a straight line)
∠h
= 16° + 107°
= 123° (Exterior angle of a triangle)
Answer(s): (a) 91°; (b) 123°