Geometry question?

One diagonal divides the rectangle into two identical right-angled triangles.
Let θ be the smallest angle in one of the right-angled triangles.

tanθ = 20/80
θ = 14°

The two diagonals divide the reactangle into four isosceles triangles.
The acute angle formed by the intersection of the diagonal is the exterior angle of the triangle with each base angle equal to 14°.
The acute angle formed by the intersection of the diagonal = 14° + 14° = 28°

13.........93 deg

sketch and label the rectangle and insert a diagonal... the length of the diagonal (d) is √(80² + 20²) = 82...5

now include the other diagonal, since this is a rectangle, the diagonals bisect each other... the acute angle formed (a) is opposite the side of length 20 and the other legs are of equal length ½d = 41...25...

now apply the law of cosines: 20² = 2(41...25)² - 2(41...25)²cos(a)

so cos(a) = (20² - 2(41...25)²)/(-2(41...25)²), solve for a (about 28°, but do not take my word for this number)

sketch and label the rectangle and insert a diagonal......... the length of the diagonal (d) is √(80² + 20²) = 82.........5

now include the other diagonal, since this is a rectangle, the diagonals bisect each other......... the acute angle formed (a) is opposite the side of length 20 and the other legs are of equal length ½d = 41.........25.........

now apply the law of cosines: 20² = 2(41.........25)² - 2(41.........25)²cos(a)

so cos(a) = (20² - 2(41.........25)²)/(-2(41.........25)²), solve for a (about 28°, but do not take my word for this number)

13.........93 deg

tan inverse of 0...........................25(20/80)

sketch and label the rectangle and insert a diagonal... the length of the diagonal (d) is √(80² + 20²) = 82...5

now include the other diagonal, since this is a rectangle, the diagonals bisect each other... the acute angle formed (a) is opposite the side of length 20 and the other legs are of equal length ½d = 41...25...

now apply the law of cosines: 20² = 2(41...25)² - 2(41...25)²cos(a)

so cos(a) = (20² - 2(41...25)²)/(-2(41...25)²), solve for a (about 28°, but do not take my word for this number)

13...93 deg

13.93 deg

Rectangle has (length,width) = (L,W) = (80,20) and diagonal length, D, = (L^2+W^2)^(1/2) = 10rt68 = 20rt17. A half-diagonal = 10rt17. Diagonals bisect each other at P, the exact center of the rectangle. Drop perpendic- ular from P to Q on a side of length 80. PQ bisects the side. PQ is10 units
in length. Isosceles triangle T with base 80 has 2 equal base angles, each = x units, where x is given by x =arctan(10/40) =arctan(1/4) =14.03624347 deg. Now the acute angle formed by the intersection of the diagonals = the sum of the interior and opposite angles in the triangle T = 2x =
28.07248694 deg.

2*arctan(20/80) ≈ 28.07°

tan inverse of 0.25(20/80)

diameter = √(80^2 + 20^2)

= 20√17

For the acute angle you have an isosceles triangle with sides, 20√17, 20√17 and 20

Let x = angle

Then using the cosine rule

20^2 = (20√17)^2 + (20√17)^2 - 2(20√17)(20√17)cos(x)

400 = 13600 - 13600cos(x)

-13200 = - 13600cos(x)

cos(x) = 33/34

x = cos^-1(33/34)

= 13.93055456

= 13.93 deg