In the figure, JKLM is a parallelogram. O is the centre of the circle. Find
- ∠JML
- ∠JKM
(a)
∠OKM = ∠KMO = 37° (Angle properties within a circle)
∠MOK
= 180° - 37° - 37°
= 106° (Angles sum of triangle)
∠OJK + ∠OKJ = ∠KOM (Exterior angle of a triangle)
∠OJK = ∠OKJ (Isosceles triangle, OJ = OK)
∠OJK
= 106° ÷ 2
= 53° (Exterior angle of a triangle)
∠JML
= 180° - 53°
= 127° (Interior angles)
(b)
∠JKM
= 180° - 53° - 37°
= 90° (Angle sum of triangles)
Answer(s): (a) 127°; (b) 90°