If one angle of a rhombus is 60° and the length of a side is 8 cm, find the lengths of its diagonals.

Each side of rhombus ABCD is 8 cm.

So, AB = BC = CD = DA = 8 cm.

Let ∠A = 60°.

In ∆ABD,

AB = AD

∠ADB = ∠ABD = x (let) (Angles opposite to equal sides are equal.)

⇒ ∠ADB + ∠ABD + ∠DAB = 180°

⇒ x + x + 60° = 180°

⇒ 2x + 60° = 180°

⇒ 2x = 180° - 60°

⇒ 2x = 120°

⇒ x = $\dfrac{120°}{2}$

⇒ x = 60°.

Since, all angles = 60°. Hence, △ABD is an equilateral triangle.

So, BD = 8 cm.

As we know, the diagonals of a rhombus bisect each other at right angles

AO = OC, BO = OD = 4cm and ∠AOB = 90°

Now, in right ∆AOB

By Pythagoras theorem,

⇒ AB² = AO² + OB²

⇒ 8² = AO² + 4²

⇒ 64 = AO² + 16

⇒ AO² = 64 – 16 = 48

⇒ AO = $\sqrt{48} = 4\sqrt{3}$ cm.

But, AC = 2AO

⇒ AC = $2 × 4\sqrt{3} = 8\sqrt{3}$ cm.

Hence, length of diagonals = 8 cm and $8\sqrt{3}$ cm.