In the figure, ABCD is a parallelogram. O is the centre of the circle. Find
- ∠ADC
- ∠ABD
(a)
∠OBD = ∠BDO = 43° (Angle properties within a circle)
∠DOB
= 180° - 43° - 43°
= 94° (Angles sum of triangle)
∠OAB + ∠OBA = ∠BOD (Exterior angle of a triangle)
∠OAB = ∠OBA (Isosceles triangle, OA = OB)
∠OAB
= 94° ÷ 2
= 47° (Exterior angle of a triangle)
∠ADC
= 180° - 47°
= 133° (Interior angles)
(b)
∠ABD
= 180° - 47° - 43°
= 90° (Angle sum of triangles)
Answer(s): (a) 133°; (b) 90°
