In the figure, ABCD and BFGH are two rhombuses. If ∠FGH = 113° and ∠ADC = 93°, calculate
- ∠GFH
- ∠GCB
- ∠HJC
(a)
∠GFH
= (180° - 113°) ÷ 2
= 33.5° (Isosceles triangle)
(b)
∠GCB = ∠ADC = 93° (Corresponding angles)
(c)
∠HJC
= 93° - 33.5°
= 59.5° (Exterior angle of a triangle)
Answer(s): (a) 33.5°; (b) 93°; (c) 59.5°