In the figure, BCDE and CGHJ are two rhombuses. If ∠GHJ = 110° and ∠BED = 102°, calculate
- ∠HGJ
- ∠HDC
- ∠JKD
(a)
∠HGJ
= (180° - 110°) ÷ 2
= 35° (Isosceles triangle)
(b)
∠HDC = ∠BED = 102° (Corresponding angles)
(c)
∠JKD
= 102° - 35°
= 67° (Exterior angle of a triangle)
Answer(s): (a) 35°; (b) 102°; (c) 67°