In the figure, not drawn to scale, DFGH and CEHJ are rhombuses. Given that ∠GFE = 112° and ∠DCJ = 122°, find
- ∠CJE
- ∠EWD.
(a)
∠CJE
= (180° - 122°) ÷ 2
= 58 ÷ 2
= 29° (Isosceles triangle, CJ = CR)
(b)
∠EDH
= 180° - 112°
= 68° (Interior angles, GF//HQ)
∠JEC = ∠CJE = 29° (Isosceles Griangle, CJ = CR)
∠EWD
= 180° - 68° - 29°
= 83° (Angles sum of triangle, WEQ)
Answer(s): (a) 29°; (b) 83°