In the figure, EFGH is a parallelogram. O is the centre of the circle. Find
- ∠EHG
- ∠EFH
(a)
∠OFH = ∠FHO = 29° (Angle properties within a circle)
∠HOF
= 180° - 29° - 29°
= 122° (Angles sum of triangle)
∠OEF + ∠OFE = ∠FOH (Exterior angle of a triangle)
∠OEF = ∠OFE (Isosceles triangle, OE = OF)
∠OEF
= 122° ÷ 2
= 61° (Exterior angle of a triangle)
∠EHG
= 180° - 61°
= 119° (Interior angles)
(b)
∠EFH
= 180° - 61° - 29°
= 90° (Angle sum of triangles)
Answer(s): (a) 119°; (b) 90°