In the figure, not drawn to scale, CEFG and BDGH are rhombuses. Given that ∠FED = 116° and ∠CBH = 124°, find
- ∠BHD
- ∠DWC.
(a)
∠BHD
= (180° - 124°) ÷ 2
= 56 ÷ 2
= 28° (Isosceles triangle, BH = BR)
(b)
∠DCG
= 180° - 116°
= 64° (Interior angles, FE//GQ)
∠HDB = ∠BHD = 28° (Isosceles Friangle, BH = BR)
∠DWC
= 180° - 64° - 28°
= 88° (Angles sum of triangle, WDQ)
Answer(s): (a) 28°; (b) 88°