In the figure, J is the centre of a semicircle and JEFG is a rhombus. Given that ∠JGH = 28°, find
- ∠q
- ∠r
(a)
∠JGH
= ∠JHG
= 28° (Isosceles triangle)
∠GJH
= 180° - 28° - 28°
= 124° (Angles sum of triangle)
∠q
= ∠GJH
= 124° (Corresponding angles)
(b)
∠r
= 180° - 124°
= 56° (Interior angles)
Answer(s): (a) 124°; (b) 56°