ABCD is a rectangle with AC =10 cm and BC=8cm. Diagonals AC and BD intersect at E and D2 =20°. Calculated the sizes of all interior angles and AD,AE and AB?

Given that ABCD is a rectangle, we know that all interior angles are right angles (90 degrees).

Since D2 = 20 degrees, we can calculate the other angles as follows:

Angle AED = 180 - D2 = 180 - 20 = 160 degrees
Angle ADE = 180 - 90 - 160 = 30 degrees
Angle BAE = 180 - 90 = 90 degrees
Angle AEB = 180 - 90 - 20 = 70 degrees
Angle EBC = 180 - 90 = 90 degrees
Angle ECB = 180 - 90 - 70 = 20 degrees

Now, we can calculate the lengths of AD, AE, and AB using trigonometry.

In triangle ADE:
sin(30) = AD / AE
AE = AD / sin(30)
AE = AD / 0.5
AE = 2 * AD

In triangle AEB:
sin(70) = AB / AE
AB = AE * sin(70)
AB = 2 * AD * sin(70)

Using the Pythagorean theorem in triangle ADE:
AD^2 + AE^2 = AC^2
AD^2 + (2AD)^2 = 10^2
AD^2 + 4AD^2 = 100
5AD^2 = 100
AD^2 = 20
AD = sqrt(20) = 2sqrt(5) cm

Now, we can calculate AE and AB:
AE = 2 * AD = 2 * 2sqrt(5) = 4sqrt(5) cm
AB = 2 * AD * sin(70) = 2 * 2sqrt(5) * sin(70) ≈ 7.66 cm

Therefore, the sizes of all interior angles are:
Angle A = 90 degrees
Angle B = 90 degrees
Angle C = 90 degrees
Angle D = 90 degrees
Angle E = 20 degrees

And the lengths of AD, AE, and AB are:
AD = 2sqrt(5) cm
AE = 4sqrt(5) cm
AB ≈ 7.66 cm