In the given figure, ABCD is a square and DEC is equilateral.
The value of x is
45°
ABCD is a square …...(given)
AB = BC = CD = DA ……. (sides of a square) ……. (i)
∠BCD = 90° ……. (angles of a square) ……… (ii)
DEC is equilateral …….. (given)
DC = CE = ED …… (sides of an equilateral triangle) ……… (iii)
Therefore, from (1) and (iii), EC =CB ….. (iv)
∠BCD = 60° ……. (angles of an equilateral triangle) ……… (v)
∠BCE = 90° + 60°
= 150°
BCE is isosceles …. (from (v))
∠CBE = ∠BEC = y
y + y + 150° = 180°
2y = 30°
y = 15°
x + y = 60°
x = 60° -15°
x= 45°
Hence (A)