Question

In the adjoining figure , ABCD is a square and CDE is an equilateral triangle. prove cone AED ,EAB ,reflex cone AEC​

Answer 1

Given ABCD is a square and CDE is a equilateral triangle.

we have to find the angles ∠AED, ∠EAB and reflex angle of ∠AEC.

Given ABCD square ∴all sides are equal and all angles are of 90°

AB=BC=CD=DA

and also CDE is an equilateral triangle

∴ CD=DE=EC and all angles are of 60°

Now, In ΔADE, ∵AD=AE ⇒ ∠DAE=∠AED

∠ADE=∠ADC-∠EDC=90°-60°=30°

By angle sum property of triangle

∠ADE+∠DAE+∠AED=180°

⇒ 30°+2∠AED=180°

⇒ 2∠AED=150°

⇒ ∠AED=75°

∠EAB=∠DAB-∠DAE=90°-75°=15°

Reflex angle

∠AEC=360°∠AED

∠DEC=360°-75°-60°=225°

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Answer 2

Step-by-step explanation:

 

ΔADE and ΔBCE are congruent due to following reasons

 

AD = BC,   DE = CE  and ADE = BCE =  (90+60)° = 150°

Hence AE = BE

 

ΔADE is isoceless triangle, because AD = DE

Hence DAE = DEA

since ADE = 150° ,  DAE = (180-150)/2 = 15°