The constant still goes to the helper
Modern compilers turn x / 3 into a multiply by a magic reciprocal and a shift. gcc 2.9 targeting Thumb does not. Every constant divisor that is not a power of two is loaded into r1 and handed to the same helper a variable divisor would use:
0 push {lr}
2 mov r1, #60
4 bl __divsi3-4
8 pop {r1}
10 bx r1
That is a division by 60, and it costs the whole call. Two consequences. Any bl __divsi3 in a listing might have a constant divisor, so check r1 before you assume the divisor is a parameter. And when a real GBA game divides by a constant in a hot loop, this is why it does not.
The divisor's value is the thing you have to recover, and materialising it follows the ladder from the arithmetic chapter. Up to 255 it is a single mov r1, #k, as above. A value that is a small number shifted left becomes mov and lsl — and the number in the mov is not the divisor, so a pair like that has to be multiplied out before it means anything. Anything else comes from the literal pool:
0 push {lr}
2 ldr r1, [pc, #8] (->12)
4 bl __divsi3-4
8 pop {r1}
10 bx r1
12 .word 257
257 needs nine significant bits, so there is no way to build it in one or two instructions, and it is parked after the code as data. The (->12) points at the row holding it.
Keep reading the (->12) rather than the #8, as you did in the arithmetic chapter. The immediate is measured from the PC, which runs four bytes ahead of the instruction and is rounded down to a multiple of four before the offset is added, so it never matches the gap you can count on the page. The arrow has already done that arithmetic.
Your target's divisor comes from the middle rung of the ladder.
Your task
Write func_080b8054 to reproduce the target assembly.