Let the center of ⊿ABC be O and connect A' O, we can see that A' O is the vertical line of ⊿ABC and connects AO.

Because the side of triangular prism ABC-a1b1c1is equal to the side length of the bottom surface, we can get AO= root 3/3*AB A`O= root 6/3*AB.

Let b be BD in parallel, which is equal to AO connecting B' D.

Therefore, the quadrilateral AODB is a parallelogram, so AB is parallel to OD, and because the triangular prism ABC-a1b1c1and A' B' is parallel to AB, so DO is parallel to A' B'.

Therefore, quadrilateral a' b' do is a parallelogram, so b' is.

At the same time, we can get AD= root 2 1/3*AB.

So AB`= root number 3*AB

So the angle between AB 1 and the bottom ABC is the angle DAB'

Then the sine value of the angle between AB 1 and the bottom ABC is equal to 2/3 of the root sign.