The sine of the acute angle of the isosceles trapezium is 0.8. find the side length

The sine of the acute angle of the isosceles trapezium is 0.8. find the side length if the base length difference is 3.

ABCD – isosceles trapezoid, AB = CD, AD – BC = 3, sinA = sinC = 0.8.
1. From the top B draw the height to AD. The height BH divides the base AD into segments, one of which is equal to the half-difference of the bases (AH), and the second is the half-sum of the bases (DH).
Consider the triangle AHB: AH and BN – legs, AB – hypotenuse.
AH = (AD – BC) / 2.
Since AD ​​- BC = 3, then:
AH = 3/2.
2. Let’s designate the side AB as x. By the Pythagorean theorem, from the triangle ANV, we find the VN:
BH = √ (AB ^ 2 – AH ^ 2);
BH = √ (x ^ 2 – (3/2) ^ 2);
BH = √ (x ^ 2 – 9/4);
BH = √ ((4x ^ 2 – 9) / 4);
BH = (√ (4x ^ 2 – 9)) / 2.
3. The sine of an acute angle in a right triangle is the ratio of the opposite leg to the hypotenuse:
sinA = BH / AB.
Substitute the values:
((√ (4x ^ 2 – 9)) / 2) / x = 8/10;
(√ (4x ^ 2 – 9)) / 2x = 4/5;
√ (4x ^ 2 – 9) = 8x / 5 (in proportion);
(√ (4x ^ 2 – 9)) ^ 2 = (8x / 5) ^ 2;
4x ^ 2 – 9 = 64x ^ 2/25;
25 (4x ^ 2 – 9) = 64x ^ 2 (in proportion);
100x ^ 2 – 225 = 64x ^ 2;
100x ^ 2 – 64x ^ 2 = 225;
36x ^ 2 = 225;
x ^ 2 = 225/36 (in proportion);
x = √ (225/36);
x = 15/6 = 5/2 = 2.5.
Thus, AB = x = 2.5.
Then: AB = CD = 2.5.
Answer: AB = CD = 2.5.