In the figure, not drawn to scale, CEFG and BDGH are rhombuses. Given that ∠FED = 115° and ∠CBH = 122°, find
- ∠BHD
- ∠DWC.
(a)
∠BHD
= (180° - 122°) ÷ 2
= 58 ÷ 2
= 29° (Isosceles triangle, BH = BR)
(b)
∠DCG
= 180° - 115°
= 65° (Interior angles, FE//GQ)
∠HDB = ∠BHD = 29° (Isosceles Friangle, BH = BR)
∠DWC
= 180° - 65° - 29°
= 86° (Angles sum of triangle, WDQ)
Answer(s): (a) 29°; (b) 86°