The angle between the diagonals of the straight line is 60 degrees and its area is -12 ^ 2 cm. Find the sides straight.

Since ABCD is a rectangle, its diagonals are equal, and at the point of their intersection they are divided in half.

The area of the rectangle is equal to: Savsd = AC ^ 2 * Sin60 / 2.

AC ^ 2 = 2 * Savsd / Sin60 = 2 * 12 / (√3 / 2) = 48 / √3 = 16 * √3 cm.

AC = 4 * 4√3

Then ОА = ОВ = АС / 2 = 2 * 4√3 cm.

The AOB triangle is equilateral, since ОА = ОВ, and the AOB angle = 60.

Then AB = ОА = ОВ = 2 * 4√3 cm.

Savsd = AB * BC.

BC = Savsd / AB = 12/2 * 4√3 = 6/4√3 cm.

Answer: The sides of the rectangle are 2 * 4√3 cm and 6 / 4√3 cm.




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