In the figure, ABCD and BFGH are two rhombuses. If ∠FGH = 105° and ∠ADC = 95°, calculate
- ∠GFH
- ∠GCB
- ∠HJC
(a)
∠GFH
= (180° - 105°) ÷ 2
= 37.5° (Isosceles triangle)
(b)
∠GCB = ∠ADC = 95° (Corresponding angles)
(c)
∠HJC
= 95° - 37.5°
= 57.5° (Exterior angle of a triangle)
Answer(s): (a) 37.5°; (b) 95°; (c) 57.5°