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 = 42° and ∠GDH = 29°. Find
- ∠ADC
- ∠DFE
(a)
∠ADH
= (180° - 42°) ÷ 2
= 69° (Isosceles triangle)
∠ADC
= 180° - 69°
= 111° (Angles on a straight line)
(b)
∠FDE
= 69° - 29°
= 40°
∠DFE
= (180° - 40°) ÷ 2
= 70 ° (Isosceles triangle)
Answer(s): (a) 111°; (b) 70°