In the figure, HJKL is a parallelogram. O is the centre of the circle. Find
- ∠HLK
- ∠HJL
(a)
∠OJL = ∠JLO = 38° (Angle properties within a circle)
∠LOJ
= 180° - 38° - 38°
= 104° (Angles sum of triangle)
∠OHJ + ∠OJH = ∠JOL (Exterior angle of a triangle)
∠OHJ = ∠OJH (Isosceles triangle, OH = OJ)
∠OHJ
= 104° ÷ 2
= 52° (Exterior angle of a triangle)
∠HLK
= 180° - 52°
= 128° (Interior angles)
(b)
∠HJL
= 180° - 52° - 38°
= 90° (Angle sum of triangles)
Answer(s): (a) 128°; (b) 90°