The angles formed when the diagonals of the rectangle intersect are ratios 2: 7.

The angles formed when the diagonals of the rectangle intersect are ratios 2: 7. Find the corners that form a diagonal with the sides of the given rectangle.

The diagonals of the rectangle are halved by the intersection point and the intersection point is halved. Thus, pairs of two equal isosceles triangles are formed.

Let a rectangle ABCD be given: AC = BD are diagonals, then △ AOB = △ COD and △ AOD = △ BOC.

1. By the condition ∠AOB: ∠BOC = 2: 7.

∠AOB + ∠BOC = 180 ° (as adjacent).

We got a system of linear equations with two variables:

∠AOB / ∠BOC = 2/7;

∠AOB + ∠BOC = 180 °.

In the first equation, we express ∠AOB:

∠AOB = (2 * ∠BOC) / 7.

Substitute the expression into the second equation:

(2 * ∠BOC) / 7 + ∠BOC = 180 °;

(2 * ∠BOC) / 7 + (7 * ∠BOC) / 7 = 180 °;

(2 * ∠BOC + 7 * ∠BOC) / 7 = 180 °;

(9 * ∠BOC) / 7 = 180 °;

∠BOC = (7 * 180 °) / 9;

∠BOC = 140 °.

Find ∠AOB:

∠AOB = (2 * 140 °) / 7 = 40 °.

2. Consider △ AOB: ∠BAO = ∠ABO = x.

∠BAO + ∠ABO + ∠AOB = 180 ° (by the theorem on the sum of the angles of a triangle);

x + x + 40 ° = 180 °;

2 * x = 180 ° – 40 °;

2 * x = 140 °;

x = 140 ° / 2;

x = 70 °.

Thus, the diagonals AC and BD form angles equal to 70 ° with the rectangle width.

3. Consider △ BOC: ∠CBO = ∠BCO = y.

∠CBO + ∠BCO + ∠BOC = 180 ° (by the theorem on the sum of the angles of a triangle);

y + y + 140 ° = 180 °;

2 * y = 180 ° – 140 °;

2 * y = 40 °;

y = 40 ° / 2;

y = 20 °.

Thus, the diagonals AC and BD form angles equal to 20 ° with the length of the rectangle.

Answer: 70 ° and 20 °.