Please, solve Q number 52

 

As, ABCD is a square and all sides of a square are equal.

AB = BC = CD = AD ----------[1]

 

Also, AEB is an equilateral triangle and all sides of an equilateral triangle are equal

 

 

AB = EA = EB [2]

 

From [1] and [2]

 

AB = BC = CD = AD = EA = EB [3]

 

 

Now,

 

 

AD = EA

 

 

⇒ ∠AED = ∠ADE [Angles opposite to equal sides are equal]

 

 

In ΔAED, By angle sum property

 

 

∠AED + ∠ADE + ∠EAD = 180°

 

 

⇒ ∠AED + ∠AED + (∠CAB - ∠EAB) = 180°

 

 

Now, ∠CAB = 90° [Angle in square] and

 

 

∠EAB = 60° [Angle in an equilateral triangle]

 

 

⇒ 2∠AED + 90° - 60° = 180°

 

 

⇒ 2∠AED = 150°

 

 

⇒ ∠AED = 75°

 

 

Similarly, ∠BEC = 75°









Regards