In the figure, not drawn to scale, BDEF and ACFG are rhombuses. Given that ∠EDC = 112° and ∠BAG = 122°, find
- ∠AGC
- ∠CWB.
(a)
∠AGC
= (180° - 122°) ÷ 2
= 58 ÷ 2
= 29° (Isosceles triangle, AG = AR)
(b)
∠CBF
= 180° - 112°
= 68° (Interior angles, ED//FQ)
∠GCA = ∠AGC = 29° (Isosceles Eriangle, AG = AR)
∠CWB
= 180° - 68° - 29°
= 83° (Angles sum of triangle, WCQ)
Answer(s): (a) 29°; (b) 83°