In the adjoining figure, equilateral △ EDC surmounts square ABCD. If ∠DEB = x°, find value of x.

Given,

△EDC is an equilateral triangle, so all sides are equal.

ED = DC = EC ……(1)

In square all sides are equal.

AB = CB = DC = AD …..(2)

From (1) and (2) we get,

DC = EC = CB

⇒ EC = CB.

In △ECB,

EC = CB

⇒ ∠BEC = ∠CBE = a (let) (Angles opposite to equal sides are equal in isosceles triangle)

From figure,

⇒ ∠C = ∠ECD + ∠DCB

⇒ ∠ECD = 60° (As each angle of a equilateral triangle = 60°)

⇒ ∠DCB = 90° (As each angle of a square = 90°)

⇒ ∠ECB = ∠ECD + ∠DCB = 60° + 90° = 150°.

In triangle BEC,

⇒ ∠BEC + ∠CBE + ∠ECB = 180°

⇒ a + a + 150° = 180°

⇒ 2a = 180° - 150°

⇒ 2a = 30°

⇒ a = 15°.

From figure,

x° = ∠DEC - ∠BEC = 60° - 15° = 45°.

Hence, x = 45.