( 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.