The correct option is B 3
In △ABC
AB2+BC2=AC2 (By Pythagoras theorem)
42+32=AC2
AC2=25
AC=5
In △ABC
∠A+∠B+∠C=180
∠A+∠C=90 (As ∠B=90).....(i)
In △ABD,
∠A+∠BDA+∠DBA=180
∠A+∠DBA=90 (As ∠BDA=90).....(i)
From (i) and (ii)
∠C=∠DBA
Similarly, ∠DBC=∠A
Now, tan∠DBC=tan∠A=PB=BCAB=34