In the figure, not drawn to scale, BDEF and ACFG are rhombuses. Given that ∠EDC = 117° and ∠BAG = 124°, find
- ∠AGC
- ∠CWB.
(a)
∠AGC
= (180° - 124°) ÷ 2
= 56 ÷ 2
= 28° (Isosceles triangle, AG = AR)
(b)
∠CBF
= 180° - 117°
= 63° (Interior angles, ED//FQ)
∠GCA = ∠AGC = 28° (Isosceles Eriangle, AG = AR)
∠CWB
= 180° - 63° - 28°
= 89° (Angles sum of triangle, WCQ)
Answer(s): (a) 28°; (b) 89°