Find the corners of a rhombus if its diagonal is 2 √3 and 2.

Given a rhombus ABCD: AC = 2√3 and BD = 2 – diagonals. The diagonals by the intersection point are halved and perpendicular to each other, then:

OA = OC = AC / 2 = 2√3 / 2 = √3;

OB = OD = BD / 2 = 2/2 = 1;

∠AOB = ∠BOC = ∠COD = ∠DOA = 90 °.

Thus, the diagonals divide the rhombus ABCD into 4 equal right-angled triangles.

1. Consider △ AOB: ∠AOB = 90 °, OA = √3 and OB = 1 – legs.

The tangent of an acute angle of a right-angled triangle is the ratio of the length of the leg opposite to the given angle to the length of the leg adjacent to the given angle.

Find the tangent ∠OAB:

tg∠OAB = OB / OA = 1 / √3 = 1 / √3 * √3 / √3 = (1 * √3) / (√3) ² = √3 / 3.

∠OAB = 30 °.

2. By the theorem on the sum of the angles of a triangle:

∠AOB + ∠OAB + ∠ABO = 180 °;

90 ° + 30 ° + ∠ABO = 180 °;

∠ABO = 180 ° – 120 °;

∠ABO = 60 °.

3. The diagonals of a rhombus are the bisectors of its angles, then:

∠A = 2 * ∠OAB = 2 * 30 ° = 60 °;

∠B = 2 * ∠ABO = 2 * 60 ° = 120 °.

Since the opposite angles of the rhombus are equal, then:

∠A = ∠C = 60 °;

∠B = ∠D = 120 °.

Answer: ∠A = 60 °, ∠B = 120 °, ∠C = 60 °, ∠D = 120 °.