In the figure, ABCD and BFGH are two rhombuses. If ∠FGH = 115° and ∠ADC = 92°, calculate
- ∠GFH
- ∠GCB
- ∠HJC
(a)
∠GFH
= (180° - 115°) ÷ 2
= 32.5° (Isosceles triangle)
(b)
∠GCB = ∠ADC = 92° (Corresponding angles)
(c)
∠HJC
= 92° - 32.5°
= 59.5° (Exterior angle of a triangle)
Answer(s): (a) 32.5°; (b) 92°; (c) 59.5°