In the figure, BCDE and CGHJ are two rhombuses. If ∠GHJ = 118° and ∠BED = 96°, calculate
- ∠HGJ
- ∠HDC
- ∠JKD
(a)
∠HGJ
= (180° - 118°) ÷ 2
= 31° (Isosceles triangle)
(b)
∠HDC = ∠BED = 96° (Corresponding angles)
(c)
∠JKD
= 96° - 31°
= 65° (Exterior angle of a triangle)
Answer(s): (a) 31°; (b) 96°; (c) 65°