In the figure, EFGH is a parallelogram. O is the centre of the circle. Find
- ∠EHG
- ∠EFH
(a)
∠OFH = ∠FHO = 34° (Angle properties within a circle)
∠HOF
= 180° - 34° - 34°
= 112° (Angles sum of triangle)
∠OEF + ∠OFE = ∠FOH (Exterior angle of a triangle)
∠OEF = ∠OFE (Isosceles triangle, OE = OF)
∠OEF
= 112° ÷ 2
= 56° (Exterior angle of a triangle)
∠EHG
= 180° - 56°
= 124° (Interior angles)
(b)
∠EFH
= 180° - 56° - 34°
= 90° (Angle sum of triangles)
Answer(s): (a) 124°; (b) 90°