In the figure, BCDE and CGHJ are two rhombuses. If ∠GHJ = 113° and ∠BED = 93°, calculate
- ∠HGJ
- ∠HDC
- ∠JKD
(a)
∠HGJ
= (180° - 113°) ÷ 2
= 33.5° (Isosceles triangle)
(b)
∠HDC = ∠BED = 93° (Corresponding angles)
(c)
∠JKD
= 93° - 33.5°
= 59.5° (Exterior angle of a triangle)
Answer(s): (a) 33.5°; (b) 93°; (c) 59.5°