In the figure, ABCD is a trapezium and ADEH is a rhombus. AD is parallel to BC and DE = DF. HDC and DGF are straight lines. ∠DAH = 51° and ∠GDH = 33°. Find
- ∠ADC
- ∠DFE
(a)
∠ADH
= (180° - 51°) ÷ 2
= 64.5° (Isosceles triangle)
∠ADC
= 180° - 64.5°
= 115.5° (Angles on a straight line)
(b)
∠FDE
= 64.5° - 33°
= 31.5°
∠DFE
= (180° - 31.5°) ÷ 2
= 74.25 ° (Isosceles triangle)
Answer(s): (a) 115.5°; (b) 74.25°