In the figure, ABCD and BFGH are two rhombuses. If ∠FGH = 119° and ∠ADC = 101°, calculate
- ∠GFH
- ∠GCB
- ∠HJC
(a)
∠GFH
= (180° - 119°) ÷ 2
= 30.5° (Isosceles triangle)
(b)
∠GCB = ∠ADC = 101° (Corresponding angles)
(c)
∠HJC
= 101° - 30.5°
= 70.5° (Exterior angle of a triangle)
Answer(s): (a) 30.5°; (b) 101°; (c) 70.5°