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° - 148°) ÷ 2
= 32° ÷ 2
= 16° (Isosceles triangle)
∠ACD
= 60° - 16°
= 44° (Equilateral triangle)
(b)
∠AED
= 60° - 16°
= 44° (Equilateral triangle)
∠ADE
= 180° - 44° - 16°
= 120° (Angles sum of triangle)
∠ADC
= 360° - 148° - 120°
= 92° (Angles at a point)
∠BAC
= ∠ACD
= 44° (Alternate angles)
∠BAE
= 60° + 44°
= 104°
Answer(s): (a) 44°; (b) 92°; (c) 104°