In the figure, EFGH is a parallelogram. O is the centre of the circle. Find
- ∠EHG
- ∠EFH
(a)
∠OFH = ∠FHO = 42° (Angle properties within a circle)
∠HOF
= 180° - 42° - 42°
= 96° (Angles sum of triangle)
∠OEF + ∠OFE = ∠FOH (Exterior angle of a triangle)
∠OEF = ∠OFE (Isosceles triangle, OE = OF)
∠OEF
= 96° ÷ 2
= 48° (Exterior angle of a triangle)
∠EHG
= 180° - 48°
= 132° (Interior angles)
(b)
∠EFH
= 180° - 48° - 42°
= 90° (Angle sum of triangles)
Answer(s): (a) 132°; (b) 90°
