ABCD is a rhombus in which ∠A = 60°. Find the ratio AC : BD.

Rhombus ABCD is shown in the figure below:

In △ABD,

⇒ AB = AD (Sides of rhombus are equal.)

⇒ ∠B = ∠D = x (let) (∵ angles opposite to equal sides are equal)

⇒ ∠A + ∠B + ∠D = 180°

⇒ 60° + x + x = 180°

⇒ 2x = 180° - 60°

⇒ 2x = 120°

⇒ x = 60°.

∴ ABD is an equilateral triangle.

So, BD = AB = AD = a (let)

Since, diagonals of rhombus bisect each other,

OB = $\dfrac{a}{2}$

In right angled triangle AOB,

AB² = AO² + OB²

a² = AO² + $\Big(\dfrac{a}{2}\Big)^2$

AO² = a² - $\dfrac{a^2}{4} = \dfrac{3a^2}{4}$

AO = $\dfrac{\sqrt{3}a}{2}$

AC = 2AO = $\sqrt{3}a$

AC : BD = $\sqrt{3}a : a = \sqrt{3} : 1$.

Hence, AC : BD = $\sqrt{3} : 1$.