In the figure, JKLM is a parallelogram. O is the centre of the circle. Find
- ∠JML
- ∠JKM
(a)
∠OKM = ∠KMO = 35° (Angle properties within a circle)
∠MOK
= 180° - 35° - 35°
= 110° (Angles sum of triangle)
∠OJK + ∠OKJ = ∠KOM (Exterior angle of a triangle)
∠OJK = ∠OKJ (Isosceles triangle, OJ = OK)
∠OJK
= 110° ÷ 2
= 55° (Exterior angle of a triangle)
∠JML
= 180° - 55°
= 125° (Interior angles)
(b)
∠JKM
= 180° - 55° - 35°
= 90° (Angle sum of triangles)
Answer(s): (a) 125°; (b) 90°