In the figure, HJKL is a parallelogram. O is the centre of the circle. Find
- ∠HLK
- ∠HJL
(a)
∠OJL = ∠JLO = 28° (Angle properties within a circle)
∠LOJ
= 180° - 28° - 28°
= 124° (Angles sum of triangle)
∠OHJ + ∠OJH = ∠JOL (Exterior angle of a triangle)
∠OHJ = ∠OJH (Isosceles triangle, OH = OJ)
∠OHJ
= 124° ÷ 2
= 62° (Exterior angle of a triangle)
∠HLK
= 180° - 62°
= 118° (Interior angles)
(b)
∠HJL
= 180° - 62° - 28°
= 90° (Angle sum of triangles)
Answer(s): (a) 118°; (b) 90°