In the given figure, ABCD is a rhombus in which ∠A = 72°. If ∠CBD = x°, find the value of x.

Given,

∠A = 72°

ABCD is a rhombus. Thus, opposite sides are parallel.

⇒ ∠A + ∠B = 180° [∵ AD ∥ BC and sum of Co-interior angles is 180°]

⇒ 72° + ∠B = 180°

⇒ ∠B = 180° - 72°

⇒ ∠B = 108°

We know that,

In a rhombus diagonals bisects the vertex angle.

⇒ ∠CBD = $\dfrac{1}{2}$ ∠B [∵ BD bisects ∠B]

⇒ ∠CBD = $\dfrac{108°}{2}$

⇒ ∠CBD = x° = 54°

⇒ x = 54.

Hence, x = 54.