In the figure, ABCD and BFGH are two rhombuses. If ∠FGH = 112° and ∠ADC = 92°, calculate
- ∠GFH
- ∠GCB
- ∠HJC
(a)
∠GFH
= (180° - 112°) ÷ 2
= 34° (Isosceles triangle)
(b)
∠GCB = ∠ADC = 92° (Corresponding angles)
(c)
∠HJC
= 92° - 34°
= 58° (Exterior angle of a triangle)
Answer(s): (a) 34°; (b) 92°; (c) 58°