Question

In the given figure ABCD is a rhombus with angle A = 67°

If DEC is an equilateral triangle, calculate:

1. ∠CBE
2. ∠DBE

Answer 1

Given that the figure ABCD is a rhombus with angle A = 67^o

In the rhombus we have

∠A = 67° = ∠C                 ...[Opposite angles]

∠A + ∠D = 180°               ...[Consecutive angles are supplementary.]

⇒ ∠D = 113°

⇒ ∠ABC = 113°

Consider ΔDBC,

DC = CB                        ...[Sides of rhombus]

So, ΔDBC is an isosceles triangle,

⇒ ∠CDB = ∠CBD

Also,

∠CDB + ∠CBD + ∠BCD = 180°

⇒ 2∠CBD = 113°

⇒ ∠CDB = ∠CBD = 56.5°      ...(i)

Consider ΔDCE,

EC = CB

So ΔBCE is an isosceles triangle

⇒ ∠CBE = ∠CEB

Also,

∠CBE + ∠CEB + ∠BCE = 180°

⇒ 2∠CBE = 53°

⇒ ∠CDE = 26.5°

From (i)

∠CBD = 56.5°

⇒ ∠CBE + ∠DBE = 56.5°

⇒ 26.5° + ∠DBE = 56.5°

⇒ ∠DBE = 56.5° - 26.5°

⇒ ∠DBE = 30°