In the figure, not drawn to scale, ACDE and ZBEF are rhombuses. Given that ∠DCB = 120° and ∠AZF = 124°, find
- ∠ZFB
- ∠BWA.
(a)
∠ZFB
= (180° - 124°) ÷ 2
= 56 ÷ 2
= 28° (Isosceles triangle, ZF = ZR)
(b)
∠BAE
= 180° - 120°
= 60° (Interior angles, DC//EQ)
∠FBZ = ∠ZFB = 28° (Isosceles Driangle, ZF = ZR)
∠BWA
= 180° - 60° - 28°
= 92° (Angles sum of triangle, WBQ)
Answer(s): (a) 28°; (b) 92°