In the figure, not drawn to scale, ACDE and ZBEF are rhombuses. Given that ∠DCB = 118° and ∠AZF = 122°, find
- ∠ZFB
- ∠BWA.
(a)
∠ZFB
= (180° - 122°) ÷ 2
= 58 ÷ 2
= 29° (Isosceles triangle, ZF = ZR)
(b)
∠BAE
= 180° - 118°
= 62° (Interior angles, DC//EQ)
∠FBZ = ∠ZFB = 29° (Isosceles Driangle, ZF = ZR)
∠BWA
= 180° - 62° - 29°
= 89° (Angles sum of triangle, WBQ)
Answer(s): (a) 29°; (b) 89°