The figure is not drawn to scale. BCEF is a square and ABF is an equilateral triangle. CDE is an isosceles triangle. AED is a straight line.
- Find ∠FAE.
- Find ∠CDE.
(a)
∠EFB = 90°
∠AFB = 60° (Equilateral triangle AFB)
∠AFE
= 90° + 60°
= 150°
∠FAE
= (180° - 150°) ÷ 2
= 30° ÷ 2
= 15° (Isosceles triangle)
(b)
∠AEC
= 90° - ∠AEF;
= 90° - 15°
= 75°
∠DEC
= 180° - ∠AEC
= 180° - 75°
= 105° (Angles on a straight line)
∠CDE
= (180° - ∠DEC) ÷ 2
= (180° - 105°) ÷ 2
= 75° ÷ 2
= 37.5° (Isosceles triangle)
Answer(s): (a) 15°; (b) 37.5°
