In the figure, BCDE and CGHJ are two rhombuses. If ∠GHJ = 115° and ∠BED = 94°, calculate
- ∠HGJ
- ∠HDC
- ∠JKD
(a)
∠HGJ
= (180° - 115°) ÷ 2
= 32.5° (Isosceles triangle)
(b)
∠HDC = ∠BED = 94° (Corresponding angles)
(c)
∠JKD
= 94° - 32.5°
= 61.5° (Exterior angle of a triangle)
Answer(s): (a) 32.5°; (b) 94°; (c) 61.5°