The figure is not drawn to scale. BCF is an equilateral triangle and ECD is an isosceles triangle. DCB and CEG are straight lines and DB//GF. ∠FCG = 90° and ∠EFC = 57°.
- Find ∠DCG.
- Find ∠CED.
- Find ∠FEG.
(a)
∠FCB = 60° (Equilateral triangle BCY)
∠FCG = 90°
∠EFC = 57°
∠DCG
= 180° - ∠FCG - ∠FCB
= 180° - 90° - 60°
= 30° (Angles on a straight line)
(b)
∠CED
= (180° - ∠DCG ÷ 2
= (180° - 30°) ÷ 2
= 150° ÷ 2
= 75° (Isosceles triangle)
(c)
∠FEC
= 180° - ∠FCG - ∠EFC
= 180° - 90° - 57°
= 33°
∠FEG
= 180° - ∠FEC
= 180° - 33°
= 147° (Angles on a straight line)
Answer(s): (a) 30°; (b) 75°; (c) 147°
