In the figure, K is the centre of a semicircle and KFGH is a rhombus. Given that ∠KHJ = 28°, find
- ∠q
- ∠r
(a)
∠KHJ
= ∠KJH
= 28° (Isosceles triangle)
∠HKJ
= 180° - 28° - 28°
= 124° (Angles sum of triangle)
∠q
= ∠HKJ
= 124° (Corresponding angles)
(b)
∠r
= 180° - 124°
= 56° (Interior angles)
Answer(s): (a) 124°; (b) 56°