In the figure, not drawn to scale, EGHJ and DFJK are rhombuses. Given that ∠HGF = 117° and ∠EDK = 124°, find
- ∠DKF
- ∠FWE.
(a)
∠DKF
= (180° - 124°) ÷ 2
= 56 ÷ 2
= 28° (Isosceles triangle, DK = DR)
(b)
∠FEJ
= 180° - 117°
= 63° (Interior angles, HG//JQ)
∠KFD = ∠DKF = 28° (Isosceles Hriangle, DK = DR)
∠FWE
= 180° - 63° - 28°
= 89° (Angles sum of triangle, WFQ)
Answer(s): (a) 28°; (b) 89°