thakurainrubisingh
Answered

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Can anyone solve this question??? plss plsss its my request to all if any one can solve this question I will mark him/ her the brainliest​

Can Anyone Solve This Question Plss Plsss Its My Request To All If Any One Can Solve This Question I Will Mark Him Her The Brainliest class=

Sagot :

thebecks

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

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