In the figure, BCDE and CGHJ are two rhombuses. If ∠GHJ = 111° and ∠BED = 94°, calculate
- ∠HGJ
- ∠HDC
- ∠JKD
(a)
∠HGJ
= (180° - 111°) ÷ 2
= 34.5° (Isosceles triangle)
(b)
∠HDC = ∠BED = 94° (Corresponding angles)
(c)
∠JKD
= 94° - 34.5°
= 59.5° (Exterior angle of a triangle)
Answer(s): (a) 34.5°; (b) 94°; (c) 59.5°