The adjoining figure shows an equilateral triangle OAB with each side = 2a units. Find the coordinates of the vertices.

Given equilateral triangle OAB.

OA = OB = AB = 2a units.

Draw AD ⊥ OB.

In an equilateral triangle, a perpendicular drawn from one of the vertices to the opposite side bisects the side.

∴ OD = $\dfrac{1}{2}$ x OB = $\dfrac{1}{2}$ x 2a = a.

In right angle triangle OAD,

⇒ OA² = OD² + AD²

⇒ (2a)² = a² + AD²

⇒ 4a² = a² + AD²

⇒ AD² = 4a² - a²

⇒ AD² = 3a²

⇒ AD = $\sqrt{3a}$ units.

⇒ AD = $\sqrt{3}$a units.

From graph,

Co-ordinates of O = (0, 0)

Co-ordinates of B = (2a, 0)

As, OD = a units and AD = $\sqrt{3}$a units.

Co-ordinates of A = (a, $\sqrt{3}$a).

Hence, co-ordinates of O = (0, 0), B = (2a, 0) and A = (a, $\sqrt{3}$a).