In the figure, not drawn to scale, BDEF and ACFG are rhombuses. Given that ∠EDC = 118° and ∠BAG = 126°, find
- ∠AGC
- ∠CWB.
(a)
∠AGC
= (180° - 126°) ÷ 2
= 54 ÷ 2
= 27° (Isosceles triangle, AG = AR)
(b)
∠CBF
= 180° - 118°
= 62° (Interior angles, ED//FQ)
∠GCA = ∠AGC = 27° (Isosceles Eriangle, AG = AR)
∠CWB
= 180° - 62° - 27°
= 91° (Angles sum of triangle, WCQ)
Answer(s): (a) 27°; (b) 91°