In the figure, ABCD and BFGH are two rhombuses. If ∠FGH = 107° and ∠ADC = 100°, calculate
- ∠GFH
- ∠GCB
- ∠HJC
(a)
∠GFH
= (180° - 107°) ÷ 2
= 36.5° (Isosceles triangle)
(b)
∠GCB = ∠ADC = 100° (Corresponding angles)
(c)
∠HJC
= 100° - 36.5°
= 63.5° (Exterior angle of a triangle)
Answer(s): (a) 36.5°; (b) 100°; (c) 63.5°