In the figure, GHJK is a parallelogram. O is the centre of the circle. Find
- ∠GKJ
- ∠GHK
(a)
∠OHK = ∠HKO = 27° (Angle properties within a circle)
∠KOH
= 180° - 27° - 27°
= 126° (Angles sum of triangle)
∠OGH + ∠OHG = ∠HOK (Exterior angle of a triangle)
∠OGH = ∠OHG (Isosceles triangle, OG = OH)
∠OGH
= 126° ÷ 2
= 63° (Exterior angle of a triangle)
∠GKJ
= 180° - 63°
= 117° (Interior angles)
(b)
∠GHK
= 180° - 63° - 27°
= 90° (Angle sum of triangles)
Answer(s): (a) 117°; (b) 90°
