In the figure, BCDE and CGHJ are two rhombuses. If ∠GHJ = 114° and ∠BED = 95°, calculate
- ∠HGJ
- ∠HDC
- ∠JKD
(a)
∠HGJ
= (180° - 114°) ÷ 2
= 33° (Isosceles triangle)
(b)
∠HDC = ∠BED = 95° (Corresponding angles)
(c)
∠JKD
= 95° - 33°
= 62° (Exterior angle of a triangle)
Answer(s): (a) 33°; (b) 95°; (c) 62°