In the given figure, ABCD is rhombus and △EDC is a equilateral. If ∠BAD = 78°, calculate :

(i) ∠CBE

(ii) ∠DBE

(i) In rhombus ABCD,

⇒ ∠C = ∠A = 78° (Opposite angles of rhombus are equal)

The sum of adjacent angles of a rhombus is always 180°.

⇒ ∠A + ∠B = 180°

⇒ 78° + ∠B = 180°

⇒ ∠B = 180° - 78° = 102°.

In △ DBC,

⇒ DC = CB (Sides of rhombus are equal in length) …..(1)

⇒ ∠CDB = ∠CBD = x (let) (In a triangle angles opposite to equal sides are equal.)

By angle sum property of triangle,

⇒ ∠CDB + ∠CBD + ∠BCD = 180°

⇒ x + x + ∠BCD = 180°

⇒ 2x + 78° = 180°

⇒ 2x = 180° - 78°

⇒ 2x = 102°

⇒ x = $\dfrac{102°}{2}$

⇒ ∠CDB = ∠CBD = 51°.

Given,

DEC is an equilateral triangle, so all the sides of triangle are equal.

∴ DC = EC …….(2)

From equations (1) and (2), we get :

⇒ CB = EC

In triangle CEB,

⇒ ∠CEB = ∠CBE = y (let) [Angles opposite to equal sides in a triangle are equal]

By angle sum property of triangle,

⇒ ∠CEB + ∠CBE + ∠ECB = 180°

⇒ y + y + (∠ECD + ∠BCD) = 180°

⇒ 2y + (60° + 78°) = 180°

⇒ 2y + 138° = 180°

⇒ 2y = 180° - 138°

⇒ 2y = 42°

⇒ y = $\dfrac{42°}{2}$

⇒ ∠CBE = 21°.

Hence, ∠CBE = 21°.

(ii) From figure,

⇒ ∠DBE = ∠CBD - ∠CBE

⇒ ∠DBE = 51° - 21° = 30°.

Hence, ∠DBE = 30°.