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 = 45° and ∠GDH = 30°. Find
- ∠ADC
- ∠DFE
(a)
∠ADH
= (180° - 45°) ÷ 2
= 67.5° (Isosceles triangle)
∠ADC
= 180° - 67.5°
= 112.5° (Angles on a straight line)
(b)
∠FDE
= 67.5° - 30°
= 37.5°
∠DFE
= (180° - 37.5°) ÷ 2
= 71.25 ° (Isosceles triangle)
Answer(s): (a) 112.5°; (b) 71.25°