The given figure shows a rhombus ABCD in which angle BCD = 80°. Find angles x and y.

It is given that in rhombus ABCD, diagonal AC and BD bisect each other at 90° and ∠ BCD = 80°.

Also, ∠BCD = ∠ BAD (opposite angles of rhombus)

⇒ ∠ BAD = 80°

Adjacent angles of a quadrilateral are supplementary.

So, ∠BCD + ∠CDA = 180°

⇒ 80° + ∠CDA = 180°

⇒ ∠CDA = 180° - 80°

⇒ ∠CDA = 100°

And, ∠ ADB = $\dfrac{1}{2}$ ∠ ADC

⇒ y = $\dfrac{1}{2}$ 100°

⇒ y = 50°

Also, ∠CDA = ∠ ABC (opposite angles of rhombus)

⇒ ∠ ABC = 100°

Diagonals bisect opposite angles.

∠ OCB or ∠ PCB = $\dfrac{1}{2}$ ∠ BCD

= $\dfrac{1}{2}\times$ 80°

= 40°

Consider triangle PCM,

Exterior angle = sum of opposite interior angles.

∠ CPD = ∠ PCM + ∠ PMC

⇒ 110° = 40° + x

⇒ x = 110° - 40°

⇒ x = 70°

Hence, the value of x = 70° and y = 50°.