In the figure, ABCD and BFGH are two rhombuses. If ∠FGH = 124° and ∠ADC = 96°, calculate
- ∠GFH
- ∠GCB
- ∠HJC
(a)
∠GFH
= (180° - 124°) ÷ 2
= 28° (Isosceles triangle)
(b)
∠GCB = ∠ADC = 96° (Corresponding angles)
(c)
∠HJC
= 96° - 28°
= 68° (Exterior angle of a triangle)
Answer(s): (a) 28°; (b) 96°; (c) 68°