In the figure, BCDE and CGHJ are two rhombuses. If ∠GHJ = 120° and ∠BED = 96°, calculate
- ∠HGJ
- ∠HDC
- ∠JKD
(a)
∠HGJ
= (180° - 120°) ÷ 2
= 30° (Isosceles triangle)
(b)
∠HDC = ∠BED = 96° (Corresponding angles)
(c)
∠JKD
= 96° - 30°
= 66° (Exterior angle of a triangle)
Answer(s): (a) 30°; (b) 96°; (c) 66°