In the figure, not drawn to scale, EGHJ and DFJK are rhombuses. Given that ∠HGF = 114° and ∠EDK = 122°, find
- ∠DKF
- ∠FWE.
(a)
∠DKF
= (180° - 122°) ÷ 2
= 58 ÷ 2
= 29° (Isosceles triangle, DK = DR)
(b)
∠FEJ
= 180° - 114°
= 66° (Interior angles, HG//JQ)
∠KFD = ∠DKF = 29° (Isosceles Hriangle, DK = DR)
∠FWE
= 180° - 66° - 29°
= 85° (Angles sum of triangle, WFQ)
Answer(s): (a) 29°; (b) 85°