In the figure, NPQR is a parallelogram. O is the centre of the circle. Find
- ∠NRQ
- ∠NPR
(a)
∠OPR = ∠PRO = 35° (Angle properties within a circle)
∠ROP
= 180° - 35° - 35°
= 110° (Angles sum of triangle)
∠ONP + ∠OPN = ∠POR (Exterior angle of a triangle)
∠ONP = ∠OPN (Isosceles triangle, ON = OP)
∠ONP
= 110° ÷ 2
= 55° (Exterior angle of a triangle)
∠NRQ
= 180° - 55°
= 125° (Interior angles)
(b)
∠NPR
= 180° - 55° - 35°
= 90° (Angle sum of triangles)
Answer(s): (a) 125°; (b) 90°
