In the figure, BCDE and CGHJ are two rhombuses. If ∠GHJ = 123° and ∠BED = 97°, calculate
- ∠HGJ
- ∠HDC
- ∠JKD
(a)
∠HGJ
= (180° - 123°) ÷ 2
= 28.5° (Isosceles triangle)
(b)
∠HDC = ∠BED = 97° (Corresponding angles)
(c)
∠JKD
= 97° - 28.5°
= 68.5° (Exterior angle of a triangle)
Answer(s): (a) 28.5°; (b) 97°; (c) 68.5°