In the figure, NP = NR, ∠PNR = 34° and ∠SQR = 15°. Find
- ∠e
- ∠f
(a)
∠NPR
= (180° - 34°) ÷ 2
= 73° (Isosceles triangle)
∠e
= 180° - 15° - 73°
= 92° (Angles sum of triangle)
(b)
∠NRQ
= 180° - 73°
= 107° (Angles on a straight line)
∠f
= 15° + 107°
= 122° (Exterior angle of a triangle)
Answer(s): (a) 92°; (b) 122°