In the figure, BCDE and CGHJ are two rhombuses. If ∠GHJ = 109° and ∠BED = 100°, calculate
- ∠HGJ
- ∠HDC
- ∠JKD
(a)
∠HGJ
= (180° - 109°) ÷ 2
= 35.5° (Isosceles triangle)
(b)
∠HDC = ∠BED = 100° (Corresponding angles)
(c)
∠JKD
= 100° - 35.5°
= 64.5° (Exterior angle of a triangle)
Answer(s): (a) 35.5°; (b) 100°; (c) 64.5°