In the given figure; ABCD is a rhombus with angle A = 67°. If DEC is an equilateral triangle, calculate :

(i) ∠CBE

(ii) ∠DBE

(i) In rhombus ABCD,

⇒ ∠C = ∠A = 67° (Opposite angles of rhombus are equal)

From figure,

⇒ ∠BCD = ∠C = 67°.

⇒ ∠A + ∠B = 180°

⇒ 67° + ∠B = 180°

⇒ ∠B = 180° - 67° = 113°.

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 + 67° = 180°

⇒ 2x = 180° - 67°

⇒ 2x = 113°

⇒ x = $\dfrac{113°}{2}$ = 56.5°

⇒ ∠CDB = ∠CBD = 56.5°

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

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

By angle sum property of triangle,

⇒ ∠CEB + ∠CBE + ∠ECB = 180°

⇒ y + y + (∠ECD + ∠BCD) = 180°

⇒ 2y + (60° + 67°) = 180°

⇒ 2y + 127° = 180°

⇒ 2y = 180° - 127°

⇒ 2y = 53°

⇒ y = $\dfrac{53°}{2}$ = 26.5°

⇒ ∠CBE = 26.5° or 26° 30'

Hence, ∠CBE = 26.5° or 26° 30'.

(ii) From figure,

⇒ ∠DBE = ∠CBD - ∠CBE

⇒ ∠DBE = 56.5° - 26.5° = 30°.

Hence, ∠DBE = 30°.