Given, ABC is triangle, where BC = 112 cm

Again, AP is perpendicular to BC

Let PC = x

then BP = 112 - x

Now, triangle APC,

       tan C = AP/PC

=> 1/cot C = AP/x

=> 1/(12/5) = AP/x

=> 5/12 = AP/x

=> AP = 5x/12  ...........1

Again, in triangle APB,

       tan B = AP/PB

=> 1/cot B = AP/(112 - x)

=> 1/(4/3) = AP/(112 - x)

=> 3/4 = AP/(112 - x)

=> AP = 3(112 - x)/4  ...........2

From equation1 and 2, we get

=> 5x/12 = 3(112 - x)/4

=> 5x = {3 * 12(112 - x)}/4

=> 5x = 3 * 3(112 - x)

=> 5x = 9(112 - x)

=> 5x = 9 * 112 - 9x

=> 5x + 9x = 9 * 112

=> 14x = 9 * 112

=> x = (9 * 112)/14

=> x = 9 * 8

=> x = 72

From, equation 1, we get

      AP = (5 * 72)/12

=> AP = 5 * 6

=> AP = 30 cm

So, the length of AP is 30 cm