Answer:
a) ∠EAB = 180° - 90° - 30° = 60°
∠EBA = 180° - 90° - 60° = 30°
a) ∠EBA = 30°
b) ∠DCA = 180° - 90° - 30° = 60°
∠EBA ≅ ∠DAC, ∠EAB ≅ ∠DCA, ∠AEB ≅ ∠CDA
ΔEBA ≅ ΔDAC because of the AAA postulate
c) EB ≅ DA, EA ≅ DC, AB ≅ CA
d) AB = CA given
sin ∠EAB = EB/AB (sin 60°) EB = (0.8660)AB
cos ∠EAB = EA/AB (cos 60°) EA = (0.5)AB
cos ∠DAC = AD/CA (cos 30°) AD = (0.8660)CA
sin ∠DAC = CD/CA (sin 30°) CD = (0.5)CA
ED = EA + AD
ED = (0.5)AB + (0.8660)CA
since AB = CA, ED = 1.366CA
since EB = (0.8660)AB and AB = CA, then EB = 0.866CA
since CD = 0.5CA,
EB + CD = 0.866CA + 0.5CA = 1.366CA
EB + CD = 1.366CA
1.366CA = 1.366CA
Proof: ED = EB + CD
= -3