The figure is not drawn to scale. CDG is an equilateral triangle and FDE is an isosceles triangle. EDC and DFH are straight lines and EC//HG. ∠GDH = 90° and ∠FGD = 58°.
- Find ∠EDH.
- Find ∠DFE.
- Find ∠GFH.
(a)
∠GDC = 60° (Equilateral triangle CDY)
∠GDH = 90°
∠FGD = 58°
∠EDH
= 180° - ∠GDH - ∠GDC
= 180° - 90° - 60°
= 30° (Angles on a straight line)
(b)
∠DFE
= (180° - ∠EDH ÷ 2
= (180° - 30°) ÷ 2
= 150° ÷ 2
= 75° (Isosceles triangle)
(c)
∠GFD
= 180° - ∠GDH - ∠FGD
= 180° - 90° - 58°
= 32°
∠GFH
= 180° - ∠GFD
= 180° - 32°
= 148° (Angles on a straight line)
Answer(s): (a) 30°; (b) 75°; (c) 148°
