In the figure, FGHJ is a parallelogram. O is the centre of the circle. Find
- ∠FJH
- ∠FGJ
(a)
∠OGJ = ∠GJO = 26° (Angle properties within a circle)
∠JOG
= 180° - 26° - 26°
= 128° (Angles sum of triangle)
∠OFG + ∠OGF = ∠GOJ (Exterior angle of a triangle)
∠OFG = ∠OGF (Isosceles triangle, OF = OG)
∠OFG
= 128° ÷ 2
= 64° (Exterior angle of a triangle)
∠FJH
= 180° - 64°
= 116° (Interior angles)
(b)
∠FGJ
= 180° - 64° - 26°
= 90° (Angle sum of triangles)
Answer(s): (a) 116°; (b) 90°
