In the figure, not drawn to scale, O is the centre of the semi circle and OEFG is a rhombus. ∠OGH = 31°. Find
- ∠q
- ∠r.
(a)
OG = OH = Radius
∠GOH
= 180° - 31° - 31°
= 118° (Isosceles triangle)
∠q = ∠GOH = 118° (Corresponding angles, EF//OG)
(b)
∠r
= 180° - 118°
= 62° (Interior angles, FG//EO)
Answer(s): (a) 118°; (b) 62°