In the figure, BCDE and CGHJ are two rhombuses. If ∠GHJ = 122° and ∠BED = 101°, calculate
- ∠HGJ
- ∠HDC
- ∠JKD
(a)
∠HGJ
= (180° - 122°) ÷ 2
= 29° (Isosceles triangle)
(b)
∠HDC = ∠BED = 101° (Corresponding angles)
(c)
∠JKD
= 101° - 29°
= 72° (Exterior angle of a triangle)
Answer(s): (a) 29°; (b) 101°; (c) 72°