PSLEABCD is a parallelogram, ACE is an equilateral triangle and CD = DE.
- Find ∠ACD.
- Find ∠CDA.
- Find ∠EAB.
(a)
CD = DE
Triangle CDE is an isosceles triangle.
∠DCE
= (180° - 150°) ÷ 2
= 30° ÷ 2
= 15° (Isosceles triangle)
∠ACD
= 60° - 15°
= 45° (Equilateral triangle)
(b)
∠AED
= 60° - 15°
= 45° (Equilateral triangle)
∠ADE
= 180° - 45° - 15°
= 120° (Angles sum of triangle)
∠ADC
= 360° - 150° - 120°
= 90° (Angles at a point)
∠BAC
= ∠ACD
= 45° (Alternate angles)
∠BAE
= 60° + 45°
= 105°
Answer(s): (a) 45°; (b) 90°; (c) 105°
