( 1)

In RT△ABC

AB = BC/sinA = 10；

AC= (ab 2-BC 2 under the radical sign) = 8.

cosA = AC/AB = 4/5；

tanB = AC/BC = 4/3；

(2)

Second, it is ok to play word games. Let's just say the front and back triangles are similar. Seconds.

Otherwise, let BC=6x, and the solution is the same as (1).

The Cos of acute angles A and B are positive because they are right triangles.

There should be a little contact with trigonometry in grade three, right? You can also kill it with trigonometry.