In the figure, MNPQ is a parallelogram. O is the centre of the circle. Find
- ∠MQP
- ∠MNQ
(a)
∠ONQ = ∠NQO = 29° (Angle properties within a circle)
∠QON
= 180° - 29° - 29°
= 122° (Angles sum of triangle)
∠OMN + ∠ONM = ∠NOQ (Exterior angle of a triangle)
∠OMN = ∠ONM (Isosceles triangle, OM = ON)
∠OMN
= 122° ÷ 2
= 61° (Exterior angle of a triangle)
∠MQP
= 180° - 61°
= 119° (Interior angles)
(b)
∠MNQ
= 180° - 61° - 29°
= 90° (Angle sum of triangles)
Answer(s): (a) 119°; (b) 90°
